Explanation

(a)
$E_{cell}$ = $E_{cell}^{{^\circ}}$ - $\frac{0.059}{n}$ log $\left\lbrack \frac{product}{reactant} \right\rbrack$
Given , $E_{{Cr}^{3 +}/Cr}^{{^\circ}}$ = -0.74 V
$E_{{Fe}^{2 +}/Fe}^{{^\circ}}$ = -0.44 V
Cr | ${Cr}^{2 +}$ (0.1 M) |${Fe}^{2 +}$| (0.01 M) / Fe
∴ ${Cr}^{3 +}\ /Cr$ is anode and${Fe}^{2 +}$ / Fe is cathode.
$E_{cell}^{{^\circ}}$ = $E_{C}^{{^\circ}}$ - $E_{A}^{{^\circ}}$
= (- 0.44) – (0.74)
= - 0.44 + 0.74
= + 0.30 V
Cell reaction is
2Cr + ${3Fe}^{2 +}$ → ${2Cr}^{3 +}$ + 3Fe
number of electrons in cell reaction = 6
$E_{cell}^{{^\circ}}$ = $E_{cell}^{{^\circ}}$ - $\frac{0.059}{n}$ log $\left\lbrack \frac{product}{reactant} \right\rbrack$
= + 0.30 V - $\frac{0.059}{6}$ log $\left\lbrack \frac{({Cr}^{3 +})^{2}}{({Fe}^{2 +})^{3}} \right\rbrack$
= 0.30 - $\frac{0.059}{6}$ log $\left\lbrack \frac{(0.1)^{2}}{(0.01)^{3}} \right\rbrack$
= 0.30 - $\frac{0.059}{6}$ log $10^{4}$
= 0.30 - $\frac{0.059}{6}$ × 0.60
= 0.30 – 5.9 × $10^{- 3}$
= 0.2941 V