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)

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Chapter19: Electrochemistry
Section: Chapter Questions
Problem 19.140QP
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Question
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+ + 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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Transcribed Image Text: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 Moving to another question will save this response. MacBook Air 80 F2 F3 d F4 F5
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