Consider a galvanic cell based upon the following half reactions: Ni 2+ + 2 e -

Question

Consider a galvanic cell based upon the following half reactions:
Ni2+ + 2 e- Ni; -0.27 V Cr3+ + 3 e- Cr; -0.73 V
Which of the following changes will increase the potential of the cell?

Adding Ni2+ ions to the nickel half reaction (assume no volume change)
Removing Cr3+ ions from solution by precipitating them out of the chromium half reaction (assume no volume change).
Adding Cr3+ ions to the chromium half reaction (assuming no volume change).
Adding equal amounts of water to both half reactions.

Explanation / Answer

Answer ) reduction reaction =>

Ni2+ + 2 e- Ni; -0.27 V

oxidation reaction

Cr ----> Cr3+ + 3 e-  -0.73 V

here more the +ve value of standard reduction potential more will be the tendency to oxidize other reduce itself.

Ecell = E cathode - E anode = (-0.27 ) - (-0.73) = 0.46 V

now the overall cell reaction will becomes:

2Cr (s) + 3Ni2+ (aq)  ----> 2Cr3+ (aq) + 3Ni    (s)

now we apply the nernst formula and we get

E= Eo - 2.303 R T /n F * log   [Ni2+]3 /[Cu3+]2

Adding Ni2+ ions, [Ni2+] increases and the value of log also increases , so the potential will decrease.

Removing  Cr3+ ions , [Cr3+] decreases and the value of log increases , so the potential will deccrease.

Adding   Cr3+ ions , [Cr3+] increases and   the value of log deccreases , so the potential will increase.

adding equal amount of water doesn't bring any change in cell potential value.

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