Question

What will be the reduction potential for the following half cell reaction at 298 K?

(Given: $\lbrack Ag^{+}\rbrack = 0.1MandEº_{cell} = + 0.80V$)

• A. $0.741V$
• B. $0.80V$
• C. $150.5KJ$
• D. $28.5kJ$

Answer 1

Answer: (A)

$0.741V$

Answer 2

$Ag^{+}e^{-} \rightarrow Ag$

$E_{cell} = Eº_{cell} - \frac{0.0591}{1}\log\frac{1}{\lbrack Ag^{+}\rbrack}$

$E_{cell} = 0.80 - \frac{0.0591}{1}\log\frac{1}{0.1} = 0.80 - 0.0591 = 0.741V$