A voltaic cell employs the following redox reaction: Sn2+( a q )+Mn( s )Sn( s )+

Question

A voltaic cell employs the following redox reaction:
Sn2+(aq)+Mn(s)Sn(s)+Mn2+(aq)
Calculate the cell potential at 25 C under each of the following conditions.

Part A

standard conditions

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Correct

Part B

[Sn2+]= 1.16×102 M ; [Mn2+]= 1.66 M .

Express your answer using two significant figures.

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Part C

[Sn2+]= 1.66 M ; [Mn2+]= 1.16×102 M .

A voltaic cell employs the following redox reaction:
Sn2+(aq)+Mn(s)Sn(s)+Mn2+(aq)
Calculate the cell potential at 25 C under each of the following conditions.

Part A

standard conditions

Ecell = 1.04   V  

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Correct

Part B

[Sn2+]= 1.16×102 M ; [Mn2+]= 1.66 M .

Express your answer using two significant figures.

Ecell=   V

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Part C

[Sn2+]= 1.66 M ; [Mn2+]= 1.16×102 M .

Explanation / Answer

Sn2+ + 2 e Sn(s) 0.13

Mn2+ + 2 e Mn(s) 1.185

if E°cell = Ered - Eox = -0.13 +1.182 = 1.052

Part B

[Sn2+]= 1.16×102 M ; [Mn2+]= 1.66 M .

so

Nernest equation

Ecell = E°cell - 0.0592/n * log(Q)

Q = [Mn2+]/[Sn2+] = 1.66 / (1.16*10^-2) = 143.103

n = 2 electrons

so

Ecell = E°cell - 0.0592/n * log(Q)

Ecell = 1.052 - 0.0592/2*log(143.103

Ecell = 0.9881

Part C

[Sn2+]= 1.66 M ; [Mn2+]= 1.16×102 M .

same case:

Ecell = E°cell - 0.0592/n * log(Q)

Q = [Mn2+]/[Sn2+] = (1.16*10^-2)/ (1.66) = 0.00698

Ecell = 1.052 - 0.0592/2*log(0.00698) = 1.11582

Ecell =1.11582 V

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