Refer to the galvanic cell below (the contents of each half-cell are written beneath each compartment): 0.10 M MnO₄^- 0.40 M Cr³⁺ 0.20 M Mn²⁺ 0.30 M Cr₂O₇²^- 0.010 M H⁺ 0.010 M H⁺ The standard reduction potentials are as follows: MnO₄^- + 8H⁺ + 5e^- $\to$ Mn²⁺ + 4H₂O,$\varepsilon$ $\circ$ = 1.506 V Cr₂O₇²^- + 14H⁺ + 6e^- $\to$ 2Cr³⁺ + 7H₂O,$\varepsilon$ $\circ$ = 1.330 V What is the value of $\varepsilon$ $\circ$ _cell?

Question

Quizplus · Accepted Answer

To calculate the standard cell potential, we need to subtract the reduction potential of the anode from the reduction potential of the cathode. The anode is the compartment with MnO4-/Mn2+. According to the table of reduction potentials, the reduction potential for MnO4-/Mn2+ is 1.51 V. The cathode is the compartment with Cr3+/Cr2+. According to the table of reduction potentials, the reduction potential for Cr3+/Cr2+ is 1.33 V. 
The standard cell potential is:
E0cell = E0cathode - E0anode
E0cell = 1.33 V - 1.51 V
E0cell = -0.18 V
Therefore, the answer is (C) -0.176 V.

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

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