Question

The saturated calomel electrode, abbreviated SCE, is often used as a reference electrode in making electrochemical measurements. The SCE is composed solution of calomel$\left({\mathrm{Hg}}_2{\mathrm{Cl}}_2\right)$. The electrolyte solution is saturated $\mathbf{KCI}$${\mathbf{E}}_{\mathbf{SCE}}$is +0.242V relative to the standard hydrogen electrode. Calculate the potential for each of the following galvanic cells containing a saturated calomel electrode and the given half-cell components as standard conditions. In each case indicate whether the SCE is the cathode or the anode. Standard reduction potentials are found in Table 11.1.

$\begin{array}{l}\mathbf{a}\mathbf{.}{\mathbf{Cu}}^{\mathbf{2}\mathbf{+}}\mathbf{+}\mathbf{2}{\mathbf{e}}^{\mathbf{-}}\mathbf{\to }\mathbf{Cu}\\ \mathbf{b}\mathbf{.}{\mathbf{Fe}}^{\mathbf{3}\mathbf{+}}\mathbf{+}{\mathbf{e}}^{\mathbf{-}}\mathbf{\to }{\mathbf{Fe}}^{\mathbf{2}\mathbf{+}}\\ \mathbf{c}\mathbf{.}\mathbf{AgCI}\mathbf{+}{\mathbf{e}}^{\mathbf{-}}\mathbf{\to }\mathbf{Ag}\mathbf{+}{\mathbf{CI}}^{\mathbf{-}}\\ \mathbf{d}\mathbf{.}{\mathbf{AI}}^{\mathbf{3}\mathbf{+}}\mathbf{3}{\mathbf{e}}^{\mathbf{-}}\mathbf{\to }\mathbf{AI}\\ \mathbf{e}\mathbf{.}{\mathbf{Ni}}^{\mathbf{2}\mathbf{+}}\mathbf{2}{\mathbf{e}}^{\mathbf{-}}\mathbf{\to }\mathbf{Ni}\end{array}$

Short answer

The cell potential$\left({\text{E}}_{\text{0}}\right)$ for the overall reaction is

a. 0.10V

b.0.53V

c.0.02V

d.1.90V

e.0.47V

Step-by-step solution

Definition of cell potential

The cell potential is the measure of the potential difference between two half cells in an electrochemical cell. The potential difference is caused by the ability of electrons to flow from one-half cells to the other.

Explanation of cell potential

The cell potential $\left({\text{E}}_{\text{0}}\right)$for the overall reaction is:

$\begin{array}{l}=0.34V-0.242V\\ =0.10V\end{array}$

Explanation of cell potential

$\begin{array}{l}=0.77V-0.242V\\ =0.53V\end{array}$

Explanation of cell potential

$\begin{array}{l}=0.24\mathrm{V}-0.242\mathrm{V}\\ =0.02\mathrm{V}\end{array}$

Explanation of cell potential

$\begin{array}{l}=0.66\mathrm{V}-0.242\mathrm{V}\\ =0.90\mathrm{V}\end{array}$

Explanation of cell potential

$\begin{array}{l}=0.24\mathrm{V}-0.23\mathrm{V}\\ =0.47\mathrm{V}\end{array}$