Question

3 Ni2+(aq) + 2 Cr(OH)3(s) + 10 OH− (aq) → 3 Ni(s) + 2 CrO42−(aq) + 8 H2O(l) ΔG∘ = +87 kJ/mol Given the standard reduction potential of the half-reaction Ni2+(aq) + 2 e− → Ni(s) E∘red = -0.28 V, calculate the standard reduction potential of the half-reaction CrO42−(aq) + 4 H2O(l) + 3 e− → Cr(OH)3(s) + 5 OH−(aq)

Answer 1

Answer : The standard reduction potential, $E^o_{(Cr^{+6}/Cr^{+3})}$ is -0.13 V.

Solution : Given,

$E^o_{(Ni^{2+}/Ni)}=-0.28V$

$\Delta G^o=+87KJ/mole=+87000J/mole$       (1 KJ = 1000 J)

The net reaction is,

$3Ni^{2+}(aq)+2Cr(OH)_3(s)+10OH^-(aq)\rightarrow 3Ni(s)+2CrO^{2-}_4(aq)+8H_2O(l)$

The half cell reactions are :

At cathode : $Ni^{2+}(aq)+2e^-\rightarrow Ni(s)$       $E^o_{(Ni^{2+}/Ni)}=-0.28V$

At anode : $CrO^{2-}_4(aq)+4H_2O(l)+3e^-\rightarrow Cr(OH)_3(s)+5OH^-(aq)$  $E^o_{(Cr^{+6}/Cr^{+3})}=?$

First we have to calculate the $E^o_{cell}$ by using formula,

$\Delta G^o=-nFE^o_{cell}$

where,

$\Delta G^o$ = Gibbs's free energy

n = number of electrons in a net chemical reaction = 6 electrons

F = Faraday constant = 96485 C

$E^o_{cell}$ = standard cell potential

Now put all the given values in this formula, we get

$+87000KJ/mole=-6\times (96485)\times E^o_{cell}\\E^o_{cell}=-0.15V$

Now we have to calculate the $E^o_{(Cr^{+6}/Cr^{+3})}$ by using formula,

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

$E^o_{cell}=E^o_{(Ni^{2+}/Ni)}-E^o_{(Cr^{+6}/Cr^{+3})}$

Now put all the given values in this formula, we get

$-0.15V=-0.28V-E^o_{(Cr^{+6}/Cr^{+3})}$

$E^o_{(Cr^{+6}/Cr^{+3})}=-0.13V$

Therefore, the standard reduction potential, $E^o_{(Cr^{+6}/Cr^{+3})}$ is -0.13 V.