Question

What will be standard cell potential of galvanic cell with the following reaction?
$2C{r}_{\left(s\right)}+3C{d}_{\left(aq\right)}^{2+}\to 2C{r}_{\left(aq\right)}^{3+}+3C{d}_{\left(s\right)}$
[Given: $E_{C{r}^{3+}/Cr}^o=-0.74$V and $E_{C{d}^{2+}/Cd}^o=-0.40$V]

• A. $0.74$V
• B. $1.14$V
• C. $0.34$V
• D. $-0.34$V

Answer 1

The correct option is C $0.34$V

$2Cr\left(s\right)+3C{d}^{2+}\left(aq\right)\to 2C{r}^{3+}\left(aq\right)+3Cd\left(s\right)$

Observing the reaction we can conclude that Cr is undergoing oxidation and acts as Anode. And $C{d}^{2+}$ is undergoing reduction and acts as cathode. Also given,

$E_{C{r}^{3+}/Cr}^0=-0.74V$ is the reduction potential of $C{r}^{3+}$

$E_{C{d}^{2+}/Cd}^0=-0.40V$ is the reduction potential of $C{d}^{2+}$

We know that,

$E_{cell}^0$=$E_{cathode}^0$-$E_{anode}^0$

Here, $E_{cell}^0$=$E_{C{d}^{2+}/Cd}^0$- $E_{C{r}^{3+}/Cr}^0$

$=-0.40-(-0.74)=0.34V$