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* (96485)* 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.