Problem: Calculate the cell potential for the following reaction as written at 25.00 C, given that [Cr2+ ] = 0.812 M and [Ni2+ ] = 0.0180 M. Standard reduction potentials can be found here.Cr(s)+Ni2+(aq)⇌Cr2+(aq)+Ni(s)

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Cr(s) → Cr2+(aq)

• Cr(s) natural elemental state oxidation number = 0
• Cr2+(aq) +2 charge oxidation number = 2
from 0 to +2 → lost 2 e- (per 1 Cr(s) atom)
oxidation number increased → lost electrons → oxidized anode


Ni2+(aq) → Ni(s)

• Ni2+(aq)  +2 charge oxidation number = +2
• Ni(s) natural elemental state oxidation number = 0
from +2 to 0 → gained 2 e- (per 1 Ni2+ ion)
oxidation number decreased reduced → cathode

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Problem Details

Calculate the cell potential for the following reaction as written at 25.00 C, given that [Cr2+ ] = 0.812 M and [Ni2+ ] = 0.0180 M. Standard reduction potentials can be found here.

Cr(s)+Ni2+(aq)⇌Cr2+(aq)+Ni(s)

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