Question

Calculate the cell potential for the following reaction as written at 25.00 °C, given that [Cr2 ] = 0.864 M and [Sn2 ] = 0.0190 M. Standard reduction potentials can be found here.

Cr(s)+Sn^2+(aq) forward and reverse arrow Cr^2+(aq)+Sn(s)

Answer 1

Lets find Eo 1st

from data table:

Eo(Cr2+/Cr(s)) = -0.91 V

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

As per given reaction/cell notation,

cathode is (Sn2+/Sn(s))

anode is (Cr2+/Cr(s))

Eocell = Eocathode - Eoanode

= (-0.13) - (-0.91)

= 0.78 V

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 {[Cr2+]^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 {[Cr2+]^1/[Sn2+]^1}

E = 0.78 - (0.0591/2) log (0.864^1/0.019^1)

E = 0.78-(4.901*10^-2)

E = 0.731 V

Answer: 0.731 V