Question

9 part 2

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

E_cell

Part B

[Sn2+]= 1.45×10⁻² M ; [Mn2+]= 1.41 M .

Express your answer using two significant figures.

E_cell

Part C

[Sn2+]= 1.41 M ; [Mn2+]= 1.45×10⁻² M .

E_cell

Answer 1

A)

Lets find Eo 1st

from data table:

Eo(Mn2+/Mn(s)) = -1.185 V

Eo(Sn2+/Sn(s)) = -0.13 V

As per given reaction/cell notation,

cathode is (Sn2+/Sn(s))

anode is (Mn2+/Mn(s))

Eocell = Eocathode - Eoanode

= (-0.13) - (-1.185)

= 1.055 V

Answer: 1.055 V

B)

Number of electron being transferred in balanced reaction is 2

So, n = 2

we have below equation to be used:

E = Eo - (2.303*RT/nF) log {[Mn2+]^1/[Sn2+]^1}

Here:

2.303*R*T/F

= 2.303*8.314*298.0/96500

= 0.0591

So, above expression becomes:

E = Eo - (0.0591/n) log {[Mn2+]^1/[Sn2+]^1}

E = 1.055 - (0.0591/2) log (1.41^1/0.0145^1)

E = 1.055-(5.877*10^-2)

E = 0.9962 V

Answer: 0.996 V

C)

E = Eo - (0.0591/n) log {[Mn2+]^1/[Sn2+]^1}

E = 1.055 - (0.0591/2) log (0.0145^1/1.41^1)

E = 1.055-(-5.877*10^-2)

E = 1.114 V

Answer: 1.11 V