Question

The reduction potential of a half-cell consisting of a Pt electrode immersed in 1.5 M Fe2+ and 0.015 M Fe3+ solution at 25⁰C is .$\left( E_{Fe^{3 +}/Fe^{2 +}}^{0} = 0.770V \right)$

• A. 0.652 V
• B. 0.88 V
• C. 0.710 V
• D. 0.850 V

Answer 1

Answer: (D)

0.850 V

Answer 2

$\text{Ag }\overset{\quad\quad}{\rightarrow}\text{ A}\text{g}^{+}\text{ + }\text{e}^{-}$

$E_{1}\text{ = }\text{E}_{\text{oxid}}\text{ + }\text{E}_{\text{calomel}}$ $= \mathrm { E } ^ { \prime } - \frac { 0.0591 } { 1 } \log \mathrm { K } _ { \text {sp } 2 } + \mathrm { E } _ { \text {calomel } }$ $E_{2} = E' - \frac{0.0591}{1}\log K_{sp2} + E_{calomel}$$\frac{K_{sp_{1}}}{K_{sp_{2}}} = 10^{3}$