Practice Exercise 19.08 What are the overall cell reaction and the standard cell

Question

Practice Exercise 19.08 What are the overall cell reaction and the standard cell potential of a galvanic cell employing the following half-reactions? Cr3+(aq) + 3e. Cr(s), Eer3+ =-0.74 V; MnO4-(aq) + 8H+(aq) + 5e- Mn2+(aq) + 4H2O, E"mn04-= 1.51 V. 5Cr(aq) + 3MnO4-(aq) + 24H+(aq) 3Mn2+(aq) + 10H20(/) + 5Cr3+(s) scr(s) + 3MnO4"(aq) + 24H +(aq) 3Mn2+(aq) + 12H20(/) + 5Cr3+(aq) 3Mn2+(aq) + 12H20() + 5Cr3+(aq) 5Cr(s) + 3MnO4-(aq) + 24H +(aq) 5Cr(s) + MnO4-(aq) + 24H+(aq) Mn2+(aq) + 12H20() + 5Cr3+(aq) What is the standard cell potential of a galvanic cell? E°cell = The number of significant digits is set to 3; the tolerance is +/-1 in the 3rd significant digit

Explanation / Answer

1)
Eo is more for 2nd reaction
So, this is cathode

To balance number of electrons we need to multiply 1st reaction by 5 and 2nd reaction by 3

5 Cr3+ (aq) + 15 e- <—> 5 Cr(s)
3 MnO4-(aq) + 24 H+(aq) + 15 e- <—> 3 Mn2+ (aq) + 12 H2O

Since 1st reaction is oxidation. It should be reversed and added to 2nd equation
Th e overall reaction will be:
3 MnO4-(aq) + 24 H+(aq) + 5 Cr(s) <—> 5 Cr3+ (aq) + 3 Mn2+ (aq) + 12 H2O

So, the answer is option 2

2)
Eo cell = Eo cathode - Eo anode
= 1.51 - (-0.74)
= 2.25 V
Answer: 2.25 V

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