Asked by Anonymous
The standard reduction potential for Cr3+(aq) is
−0.74 V.
The half-reaction for the reduction of Cr3+(aq) is the following.
Cr3+(aq) + 3 e− → Cr(s)
The standard reduction potential for Ni2+(aq) is
−0.26 V.
The half-reaction for the reduction of Ni2+(aq) is the following.
Ni2+(aq) + 2 e− → Ni(s)
Using this information, calculate E⁰cell for the voltaic cell powered by the following spontaneous redox reaction.
3 Ni2+(aq) + 2 Cr(s) → 2 Cr3+(aq) + 3 Ni(s)
−0.74 V.
The half-reaction for the reduction of Cr3+(aq) is the following.
Cr3+(aq) + 3 e− → Cr(s)
The standard reduction potential for Ni2+(aq) is
−0.26 V.
The half-reaction for the reduction of Ni2+(aq) is the following.
Ni2+(aq) + 2 e− → Ni(s)
Using this information, calculate E⁰cell for the voltaic cell powered by the following spontaneous redox reaction.
3 Ni2+(aq) + 2 Cr(s) → 2 Cr3+(aq) + 3 Ni(s)
Answers
Answered by
DrBob222
3Ni^2+ + 2e ==> 3Ni(s) Eo red = - -0.26
2Cr(s) ==> 2Cr^3+ + 3e Eo ox = +0.74
.................Eocell = Eo red + Eo ox = -0.26+ 0.74 = ?
2Cr(s) ==> 2Cr^3+ + 3e Eo ox = +0.74
.................Eocell = Eo red + Eo ox = -0.26+ 0.74 = ?
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