A galvanic cell is prepared according to the redox reaction shown below. A voltmeter connected to this galvanic cell under standard conditions reads 0.25 V, and delta G of reaction= −44 kJ/mol.

Question

A galvanic cell is prepared according to the redox reaction shown below. A voltmeter connected to this galvanic cell under standard conditions reads 0.25 V, and delta G of reaction= −44 kJ/mol. If the concentration of Ni2+ is increased to 2.0 M, then E of cell will ______ and ΔGrxn will be _________ than ΔG∘rxn.

Ni(s) + 2H+(aq) ⟶ Ni2+ + H2(g)

decrease, more negative

 increase, more negative

  increase, less negative

 decrease, less negative

Answer 1

Looking at the Nernst equation, we see that..

Ecell = Eºcell - RT/nF ln Q and at STP Ecell = Eºcell - 0.0592/n log Q

If [Ni²⁺] increases, then Q will increase since Q = [Ni²⁺]/[H⁺]

If Q increases, log Q will increase

If log Q increases Ecell will decrease

∆Gº = -nFEºcell

∆G = -nFEcell

n and F are constant so ∆G will vary with Ecell

From above, increase in [Ni²⁺] will results in a decrease in Ecell

A decrease in Ecell will result in a less negative value for ∆G

So, ∆G will have a less negative value than ∆Gº

So, I'm guessing the answer would be

decrease, less negative