Question

Calculate the emf of the following cell at $298\text{K}$.
$Zn\left(s\right)\mid Z{n}^{2+}(aq,0.1M)\parallel A{g}^{+}(aq,0.01M)\mid Ag\left(s\right)$
Given:
$E_{Z{n}^{2+}/Zn}^0=-0.76V$
$E_{A{g}^{+}/Ag}^0=+0.80V$

• A. $E_{cell}=1.472V$
• B. $E_{cell}=-1.472V$
• C. $E_{cell}=+1.732V$
• D. $E_{cell}=-1.732V$

Answer 1

The correct option is A $E_{cell}=1.472V$

From the given cell representation,
$A{g}^{+}/Ag$ couple act as cathode
$Z{n}^{2+}/Zn$ couple act as anode

$E_{cell}^0=E_{cathode}^0-E_{anode}^0$
$E_{cell}^0=0.80-(-0.76)$
$E_{cell}^0=1.56V$
The given cell reaction is,
$Zn\left(s\right)+2A{g}^{+}\left(aq\right)\to Z{n}^{2+}\left(aq\right)+2Ag$
$n=2$
By Nernst equation,
$E_{cell}=E_{cell}^o-\frac{0.0591}{n}log\frac{\left[Z{n}^{2+}\right]}{{}^{{\left[A{g}^{+}\right]}^2}}$

$E_{cell}=E_{cell}^o+\frac{0.0591}{n}log\frac{{}^{{\left[A{g}^{+}\right]}^2}}{\left[Z{n}^{2+}\right]}$

$E_{cell}=1.56+\frac{0.0591}{2}log\frac{{\left[0.01\right]}^2}{\left[0.1\right]}$

$E_{cell}=1.56+\frac{0.0591}{2}log(1\times {10}^{-3})$

$E_{cell}=1.56-\frac{0.0591}{2}\times 3$
$E_{cell}=1.56-0.088=1.472V$