Answer : The standard reduction potential, [tex]E^o_{(Cr^{+6}/Cr^{+3})}[/tex] is -0.13 V.
Solution : Given,
[tex]E^o_{(Ni^{2+}/Ni)}=-0.28V[/tex]
[tex]\Delta G^o=+87KJ/mole=+87000J/mole[/tex] (1 KJ = 1000 J)
The net reaction is,
[tex]3Ni^{2+}(aq)+2Cr(OH)_3(s)+10OH^-(aq)\rightarrow 3Ni(s)+2CrO^{2-}_4(aq)+8H_2O(l)[/tex]
The half cell reactions are :
At cathode : [tex]Ni^{2+}(aq)+2e^-\rightarrow Ni(s)[/tex] [tex]E^o_{(Ni^{2+}/Ni)}=-0.28V[/tex]
At anode : [tex]CrO^{2-}_4(aq)+4H_2O(l)+3e^-\rightarrow Cr(OH)_3(s)+5OH^-(aq)[/tex] [tex]E^o_{(Cr^{+6}/Cr^{+3})}=?[/tex]
First we have to calculate the [tex]E^o_{cell}[/tex] by using formula,
[tex]\Delta G^o=-nFE^o_{cell}[/tex]
where,
[tex]\Delta G^o[/tex] = Gibbs's free energy
n = number of electrons in a net chemical reaction = 6 electrons
F = Faraday constant = 96485 C
[tex]E^o_{cell}[/tex] = standard cell potential
Now put all the given values in this formula, we get
[tex]+87000KJ/mole=-6\times (96485)\times E^o_{cell}\\E^o_{cell}=-0.15V[/tex]
Now we have to calculate the [tex]E^o_{(Cr^{+6}/Cr^{+3})}[/tex] by using formula,
[tex]E^o_{cell}=E^o_{cathode}-E^o_{anode}[/tex]
[tex]E^o_{cell}=E^o_{(Ni^{2+}/Ni)}-E^o_{(Cr^{+6}/Cr^{+3})}[/tex]
Now put all the given values in this formula, we get
[tex]-0.15V=-0.28V-E^o_{(Cr^{+6}/Cr^{+3})}[/tex]
[tex]E^o_{(Cr^{+6}/Cr^{+3})}=-0.13V[/tex]
Therefore, the standard reduction potential, [tex]E^o_{(Cr^{+6}/Cr^{+3})}[/tex] is -0.13 V.
