Question

A voltaic cell is constructed that uses the following reaction and operates at \(298 \mathrm{K} :\) $$ \mathrm{Zn}(s)+\mathrm{Ni}^{2+}(a q) \longrightarrow \mathrm{Zn}^{2+}(a q)+\mathrm{Ni}(s) $$ (a) What is the emf of this cell under standard conditions? (b) What is the emf of this cell when \(\left[\mathrm{Ni}^{2+}\right]=3.00 M\) and \(\left[\mathrm{Zn}^{2+}\right]=0.100 \mathrm{M} ?(\mathbf{c})\) What is the emf of the cell when \(\left[\mathrm{Ni}^{2+}\right]=0.200 \mathrm{M}\) and \(\left[\mathrm{Zn}^{2+}\right]=0.900 \mathrm{M} ?\)

Short answer

(a) The emf of the cell under standard conditions is 0.53 V. (b) The emf of the cell when \([Ni^{2+}] = 3.00 \, M \) and \([Zn^{2+}] = 0.100 \, M \) is 0.56 V. (c) The emf of the cell when \([Ni^{2+}] = 0.200 \, M \) and \([Zn^{2+}] = 0.900 \, M \) is 0.42 V.

Step-by-step solution

Look up standard reduction potentials

Check the standard reduction potentials of Zn²⁺ and Ni²⁺. They are: Zn²⁺/Zn: -0.76 V Ni²⁺/Ni: -0.23 V Step 2: Calculate the standard cell potential

Find the value of the cell potential at standard conditions.

The overall cell potential at standard condition is given by $$E°_{cell} = E°_{cathode} - E°_{anode} $$ $$E°_{cell} = (-0.23 \, \text{V}) - (-0.76 \, \text{V}) = 0.53 \, \text{V}$$ (a) The emf of the cell under standard conditions is 0.53 V. Step 3: Apply the Nernst Equation for given concentrations

Calculate the emf for given concentrations using the Nernst equation.

For part (b), we are given concentrations of \([\mathrm{Ni}^{2+}] = 3.00 \, M \) and \([\mathrm{Zn}^{2+}] = 0.100 \, M \). The Nernst equation is used for this calculation: $$E = E° - \dfrac{RT}{nF}\ln Q$$ First, calculate the reaction quotient Q for the given concentrations: $$Q = \dfrac{\left[\mathrm{Zn}^{2+}\right]}{\left[\mathrm{Ni}^{2+}\right]} = \dfrac{0.100}{3.00}$$ Now, substitute the values into the Nernst equation, with n = 2 for the exchange of 2 electrons in the reaction. $$E = 0.53 - \dfrac{(8.314)(298)}{(2)(96485)}\ln \left(\dfrac{0.100}{3.00}\right)$$ $$E = 0.56 \, \text{V}$$ (b) The emf of the cell when \([Ni^{2+}] = 3.00 \, M \) and \([Zn^{2+}] = 0.100 \, M \) is 0.56 V. For part (c), we are given concentrations of \([Ni^{2+}] = 0.200 \, M \) and \([Zn^{2+}] = 0.900 \, M \). Reapply the Nernst equation with the new given concentrations. First, calculate the reaction quotient Q for the given concentrations: $$Q = \dfrac{\left[\mathrm{Zn}^{2+}\right]}{\left[\mathrm{Ni}^{2+}\right]} = \dfrac{0.900}{0.200}$$ Next, substitute the values into the Nernst equation: $$E = 0.53 - \dfrac{(8.314)(298)}{(2)(96485)}\ln \left(\dfrac{0.900}{0.200}\right)$$ $$E = 0.42 \, \text{V}$$ (c) The emf of the cell when \([Ni^{2+}] = 0.200 \, M \) and \([Zn^{2+}] = 0.900 \, M \) is 0.42 V.

Key concepts

Understanding Standard Reduction Potentials

Grasping the concept of standard reduction potentials is crucial when we delve into the workings of electrochemical cells like the voltaic cell in our exercise. These potentials, measured in volts, indicate the tendency of a chemical species to acquire electrons and thereby be reduced. Each half-cell reaction comes with its own standard reduction potential value, which has been determined under standard conditions: a pressure of 1 atm, temperature of 25°C (298K), and 1M concentration for all solutions.

The standard reduction potentials for zinc and nickel ions are essential for determining the cell potential. Positive values suggest a greater tendency for reduction, while negative values indicate less inclination. Recall that we had the values for Zn²⁺ and Ni²⁺ which show that Ni²⁺ has a higher tendency to gain electrons compared to Zn²⁺. Understanding these principles allows us to predict which direction electrons will flow in the cell and ultimately calculate the voltage that the cell can produce.

Determining Cell Potential

The cell potential, also known as electromotive force (EMF), is what drives the electric current in a voltaic cell. It's a measure of the energy per unit charge that is available from the redox reaction occurring within the cell. For our voltaic cell example, the EMF under standard conditions can be calculated using the standard reduction potentials of the half-reactions.

We apply the formula: \(E°_{cell} = E°_{cathode} - E°_{anode}\). Here, the cathode is the electrode where reduction takes place and the anode is where oxidation occurs. Positive cell potential means the reaction is spontaneous, which is the case with our voltaic cell, resulting in a 0.53 V under standard conditions. Handling these calculations is key to understanding the potential energy changes during redox reactions.

The Nernst Equation and Its Applications

Moving beyond standard conditions, the Nernst equation allows us to estimate the cell potential for non-standard conditions—where temperatures, pressures, or concentrations differ from the standard values. This equation integrates the standard cell potential with the reaction quotient (Q), temperature (T), the number of moles of electrons transferred in the reaction (n), the gas constant (R), and the Faraday constant (F) to calculate the actual cell potential.

In simple terms, the Nernst equation tells us how sensitive a cell's EMF is to changes in substance concentrations. For instance, changing the concentrations of either Zn²⁺ or Ni²⁺ ions will shift the cell potential. In our exercise, we used this equation to find that varying concentrations of Ni²⁺ and Zn²⁺ ions resulted in different EMFs, clearly demonstrating the direct influence concentrations have on the cell's potential.

Interpreting the Reaction Quotient (Q)

The reaction quotient (Q) is calculated like the equilibrium constant (K), but for a reaction that has not necessarily reached equilibrium. Q represents the ratio of product activities to reactant activities at any point in time during a reaction. For the voltaic cell we analyzed, Q is the ratio of the concentrations of Zn²⁺ ions to Ni²⁺ ions.

When we introduce the reaction quotient into the Nernst equation, we are accounting for how the reaction's position (relative to equilibrium) affects the cell potential. The larger the product of the reaction quotient, the smaller the cell's EMF becomes—we saw this when the EMF decreased as the concentration of Zn²⁺ increased relative to Ni²⁺. Mastery of the reaction quotient is fundamental, as it unveils the dynamic nature of cell potential based on the shifting landscapes of reactants and products.