Question

What is the emf at 25⁰ C for the cell,

$Ag\left\lbrack \begin{aligned} & AgBr(S),Br \\ & a = 0.34 \\ & \end{aligned} \right\rbrack\left\lbrack \begin{aligned} & Fe^{3 +},Fe^{2 +} \\ & a = 0.1a = 0.02 \end{aligned} \right\rbrack Pt$The

standard reduction potentials for the half-reactions

$\text{AgBr + e}\text{¯}\text{ } \rightarrow \text{Ag + Br}\text{¯}\text{ and F}\text{e}^{\text{3+}}\text{ + e}\text{¯} \rightarrow \text{F}\text{e}^{\text{2+ }}$are

+ 0.0713 V and + 0.770 V respectively.

• A. 0.474 volt
• B. 0.529 volt
• C. 0.356 volt
• D. 0.713 volt

Answer 1

Answer: (D)

0.713 volt

Answer 2

Ecell = (0.77 – 0.0713) –$\frac{0.059}{1}$ log $\frac{0.02}{0.1 \times 0.34}$

= 0.713 volt.