Refer to the galvanic cell below (the contents of each half-cell are written beneath each compartment): 0.10 M MnO 4 - 0.40 M Cr 3+
0.20 M Mn 2+ 0.30 M Cr 2 O 7 2 -
0.010 M H + 0.010 M H +
The standard reduction potentials are as follows:
MnO 4 - + 8H + + 5e - $\to$ Mn 2+ + 4H 2 O,$\varepsilon$ $\circ$ = 1.51 V
Cr 2 O 7 2 - + 14H + + 6e - $\to$2Cr 3+ + 7H 2 O,$\varepsilon$ $\circ$ =1.33 V
-When current is allowed to flow,which species is oxidized?

Question

Refer to the galvanic cell below (the contents of each half-cell are written beneath each compartment):  0.10 M MnO 4 - 0.40 M Cr 3+ 
0.20 M Mn 2+ 0.30 M Cr 2 O 7 2 - 
0.010 M H + 0.010 M H + 
The standard reduction potentials are as follows:
MnO 4 - + 8H + + 5e - $\to$ Mn 2+ + 4H 2 O,$\varepsilon$ $\circ$ = 1.51 V
Cr 2 O 7 2 - + 14H + + 6e - $\to$2Cr 3+ + 7H 2 O,$\varepsilon$ $\circ$ =1.33 V
-When current is allowed to flow,which species is oxidized?

Quizplus · Accepted Answer

In a galvanic cell, oxidation occurs at the anode, while reduction occurs at the cathode. The species with the highest reduction potential will be reduced at the cathode, while the species with the lowest reduction potential will be oxidized at the anode. Comparing the reduction potentials given, Cr³⁺ has the lowest reduction potential (1.33 V), indicating that it has the highest tendency to be oxidized. Therefore, the species that is oxidized in this galvanic cell is choice B, Cr³⁺.

Refer to the Galvanic Cell Below (The Contents of Each →\to→

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