16 Consider the following reaction with manganese and vanadium:2 Mn²+ (aq) + √²+ (aq)->>>2 Mn3+ (aq) + V(s)When this reaction reaches equilibrium, what will be the cell potential (V)?Mn³+ + eMn2+E = +0.32->>V2+ + 2e → VE = -0.49-(the answer should be entered with 3 DIGITS; do not enter units; give answer inand 123. and 12.3)LMoving to another question will save this response.MacBook Air80F2F3dF4F5

Ecell =Ecathode- Eanode Cathode is Vanadium anode is Manganese

Answered: Mn³+ + e- → Mn²+ E = +0.32 v2+ + 2e → V…

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Mn³+ + e- → Mn²+ E = +0.32 v2+ + 2e → V E = -0.49 - (the answer should be entered with 3 DIGITS; and 123. and 12.3)

Chemistry: The Molecular Science

5th Edition

ISBN:9781285199047

Author:John W. Moore, Conrad L. Stanitski

Publisher:John W. Moore, Conrad L. Stanitski

Chapter17: Electrochemistry And Its Applications

Section: Chapter Questions

Problem 38QRT

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Question

Consider the following reaction with manganese and vanadium:
2 Mn²+ (aq) + √²+ (aq)
->>>
2 Mn3+ (aq) + V(s)
When this reaction reaches equilibrium, what will be the cell potential (V)?
Mn³+ + e
Mn2+
E = +0.32
->>
V2+ + 2e → V
E = -0.49
-
(the answer should be entered with 3 DIGITS; do not enter units; give answer in
and 123. and 12.3)
L
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