Question

A galvanic cell is constructed that has a zinc anode immersed in a $\mathbf{\text{Zn}}{\left({\text{NO}}_{\text{3}}\right)}_{\mathbf{\text{2}}}$solution and a platinum cathode immersed in an NaCl solution equilibrated with ${\mathbf{\text{Cl}}}_{\mathbf{\text{2}}}\mathbf{\text{(g)}}$at 1 atm and 25°C. A salt bridge connects the two half-cells.

1. Write a balanced equation for the cell reaction.
2. A steady current of 0.800 A is observed to flow for a period of 25.0 minutes. How much charge passes through the circuit during this time? How many moles of electrons is this charge equivalent to?
3. Calculate the change in mass of the zinc electrode.
4. Calculate the volume of gaseous chlorine generated or consumed as a result of the reaction.

Short answer

1. Balanced equation for the cell reaction is $Zn\left(s\right)+C{l}_2\left(g\right)\to Z{n}^{2+}\left(aq\right)+2C{l}^{-}\left(aq\right)$.
2. Charge passes through the circuit is $Q=1.20\times {10}^3\text{C}$. The number of moles is $1.24\times {10}^{-2}\text{mol}$which is equivalent to charge Q.
3. The change in mass of the zinc electrode is $0.407g$.
4. The volume of gaseous chlorine generated as a result of the reaction is$0.139L$ .

Step-by-step solution

Electrochemical cells

Electrochemical reactions are made up of two half-reactions, the anode electrode and the cathode electrode, which are combined to form the entire cell reaction.

Number of moles calculated using formula: $\mathbf{\text{n=}}\frac{\mathbf{\text{aquantityofelectricity}}}{\mathbf{\text{chargefor4molesofelectrons}}}$

The mass of a substance produced during electrolysis can be calculated from the charge transferred, the faraday and the relative atomic massdata-custom-editor="chemistry" $\left(A_r\right)$ or relative formula mass$\left(M_r\right)$ of the substance.

Understanding given parameters

Zinc anode immersed in a$Zn{\left(N{O}_3\right)}_2$ solution and platinum cathode immersed in an NaCl solution equilibrated with$C{l}_2\left(g\right)$ at 1 atm and 25°C.

Current (I) is 0.800 A and time is 25 minutes.

Writing balanced equation for cell reaction

(a)

For oxidation of Zn, anode half-cell reaction can be written as,

$\text{Zn}\left(s\right)\to {\text{Zn}}^{2+}\left(aq\right)+2e^{-}$

For reduction of $C{l}_2$, cathode half-cell reaction can be written as,

${\text{Cl}}_2\left(g\right)+2e^{-}\to 2{\text{Cl}}^{-}\left(aq\right)$

Overall reaction now written as adding both half-reactions,

$\text{Zn}\left(s\right)+{\text{Cl}}_2\left(g\right)\to {\text{Zn}}^{2+}\left(aq\right)+2{\text{Cl}}^{-}\left(aq\right)$

Therefore, balanced equation for the overall reaction in given galvanic cell is,

$\text{Zn}\left(s\right)+{\text{Cl}}_2\left(g\right)\to {\text{Zn}}^{2+}\left(aq\right)+2{\text{Cl}}^{-}\left(aq\right)$

Computing number of moles of electrons

(b)

Charge passes through circuit given as$Q=I\cdot t$

$\because I=0.800A=0.800C{s}^{-1}$

And$t=25\min =25\times 60s=1500s$

$\begin{array}{c}Q=I\cdot t\\ =0.800{\text{Cs}}^{-1}\times 1500\text{s}\\ =1200\text{C}\\ =1.20\times {10}^3\text{C}\end{array}$

Therefore, Charge passes through the circuit is $Q=1.20\times {10}^3\text{C}$.

Number of moles calculate as:

$\begin{array}{c}n=\frac{aquantityofelectricity}{\text{charge}for4molesofelectrons}\\ =\frac{0.800{\text{Cs}}^{-1}\times 1500\text{s}}{96,485{\text{Cmol}}^{-1}}\\ =1.24\times {10}^{-2}\text{mol}\end{array}$

Therefore, number of moles equivalent to charge Q is $1.24\times {10}^{-2}\text{mol}$.

Calculating change in mass of zinc electrode

(c)

Relative formula mass of zinc$M_r\left(Zn\right)=65.38{\text{gmol}}^{-1}$

Change in mass of zinc electrode formulated as,

$\begin{array}{c}\text{m}\left(\text{Zn}\right)=\text{n}\times \frac{1\text{molZn}}{2\text{mol}}\times {\text{M}}_r\left(\text{Zn}\right)\\ =1.24\times {10}^{-2}\text{mol}\times \frac{1\text{molZn}}{2\text{mol}}\times 65.38{\text{gmol}}^{-1}\\ =0.407g\end{array}$

Therefore, the change in mass of the zinc electrode is $0.407g$.

Calculating volume of chlorine

(d)

Volume of 1 mol of gas is $22.4{\text{Lmol}}^{-1}$.

Volume of a chlorine gas generated as a result of reaction is formulated as,

$\begin{array}{c}\text{V}\left({\text{Cl}}_2\right)=n\times \frac{1{\text{molCl}}_2}{2\text{mol}}\times \text{V}(1\text{mol})\\ =1.24\times {10}^{-2}\text{mol}\times \frac{1{\text{molCl}}_2}{2\text{mol}}\times 22.4{\text{Lmol}}^{-1}\\ =0.139\text{L}\end{array}$

Therefore, the volume of gaseous chlorine generated as a result of the reaction is $0.139L$.