Question

A galvanic cell is formed by immersing $Ag\left(s\right)$ in $AgN{O}_3(aq,0.1M)$ and $Cu\left(s\right)$ in $CuS{O}_4(aq,4M)$ at $298\text{K}$
Calculate the EMF of the cell.
Given:
$E_{A{g}^{+}/Ag}^0=0.80V$
$E_{C{u}^{2+}/Cu}^0=0.34V$

• A. $E_{cell}=0.301V$
• B. $E_{cell}=0.383V$
• C. $E_{cell}=0.45V$
• D. $E_{cell}=1.04V$

Answer 1

The correct option is B $E_{cell}=0.383V$

Standard reduction potential of $A{g}^{+}/Ag$ couple has value higher value, hence, it will act as cathode and $C{u}^{2+}/Cu$ will act as anode.

Thus, the cell representation will be,
$Cu\left(s\right)\mid C{u}^{2+}(aq,4.0M)\parallel A{g}^{+}(aq,0.1M)\mid Ag\left(s\right)$
Cell reaction is
$Cu+2A{g}^{+}\to C{u}^{2+}+2Ag$
Here $\text{n = 2}$;
$E=E_{cell}^0-\frac{0.059}{2}log\frac{\left[C{u}^{2+}\right]}{[A{g}^{+}{]}^2}$
$E_{cell}^0=E_{cathode}^0-E_{anode}^0$
$E_{cell}^0=0.80-0.34$
$E_{cell}^0=0.46V$

By Nernst equation,
$E_{cell}=0.46-\frac{0.059}{2}log\frac{\left[4.0\right]}{{\left[0.1\right]}^2}$
$E_{cell}=0.46-\frac{0.059}{2}log(4\times {10}^2)$
$E_{cell}=0.46-\frac{0.059}{2}\times 0.602-0.059$
$E_{cell}=0.46-0.018-0.059$
$E_{cell}=0.46-0.077$
$E_{cell}=0.383V$