Question

Cell equation : $A + 2B^+$ $\to$ $A^{2+} + 2B$
$A^{2+} + 2e^- \to A; E^\circ = +0.34 \,V$
and $log_{10} K \,= \,15.6 \,at \,300 \,K$ for cell reactions.
Find $E^\circ$ for $B^+ + e^- \to B$.
[Given : $\frac{2.303\, RT}{nF}=\, 0.059\, at\, 300\, K$]

• A. 0.80 V
• B. 1.26 V
• C. -0.54 V
• D. +0.94 V

Answer 1

Answer: (A)

$E^{\circ}_{cell}=\frac{0.059}{2}\,log\,K$
$E^{\circ}_{B^{+}\,/\,B}\,\,-E^{\circ}_{A^{2+}\,/\,A}=\frac{0.059}{2}\,log\,K$
$E^{\circ}_{B^{+}\,/\,B}\,\,-0.34\,V=\frac{0.059}{2}\times15.6$
$E^{\circ}_{B^{+}\,/\,B}=0.46+0.34=0.80\,V$