Question

Calculate the standard cell potentials of galvanic cells in which the following reactions take place: 
 (i)  \(2Cr_{(S)}+3Cd^{2+}_{(aq)} \rightarrow 2Cr^{3+}_{(aq)}+3Cd\)
 (ii) \(Fe^{2+}_{(aq)}+Ag^+_{(aq)} \rightarrow Fe^{3+}_{(Aq)}+Ag(s)\)
Calculate the \(\triangle_rG^\ominus\) and equilibrium constant of the reactions.

Answer 1

(i) \(E^{\ominus}_{Cr^{3+} / Cr} = 0.74 V\)

\(E^\ominus_{Cd^{2+}/ Cd} = -0.40 V\)

The galvanic cell of the given reaction is depicted as:

\(Cr_{(s)}|Cr^{3+}_{(aq)}||Cd^{2+}_{(aq)}|Cd_{(s)}\)

Now, the standard cell potential is

\(E^\ominus_{cell}\)= \(E^{\ominus}_{R}\)-\(E^\ominus_L\)

=-0.40-(-0.74)

=+0.34 V

\(\triangle_tG^\ominus = -nFE^\ominus_{cell}\)

in the given equation,

n=6

F= 96487 C mol^-1

\(E^\ominus_{cell}\) = +0.34 V

Then, \(\triangle_tG^\ominus\) = -6 * 96487 C mol^-1  \(\times\) 0.34 V

= - 196833.48 CV mol^-1

= - 196833.48 j mol^-1

= - 196.83 kj mol^-1

Again

\(\triangle_tG^\ominus\) = -RT in K

\(\Rightarrow \triangle_tG^\ominus\)  = - 2.303 \(R\)\(\Tau\) in K

\(\Rightarrow\) log K = - \(\frac{\triangle_tG^\ominus}{2.303RT}\)

= \(\frac{-196.83 \times 10^3}{2.303 \times 8.314 \times 298}\)

= 34.496 

∴ K = antilog (34.496) = 3.13 × 10³⁴

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(ii) \(E^\ominus\) Fe³⁺/Fe²⁺ = 0.77 V

\(E^\ominus\) Ag⁺/Ag = -0.80 V

The galvanic cell of the given reaction is depicted as:

Fe²⁺ _(aq) | Fe³⁺_(aq)||Ag⁺_(aq)|Ag_(s)

Now, the standard cell potential is

\(E^\ominus_{cell}\) = \(E^{\ominus}_{R}\)- \(E^\ominus_L\)

= 0.80-0.77

=0.03 V

Here, n = 1.

Then, \(\triangle_tG^\ominus = -nFE^\ominus_{cell}\)

= - 1 × 96487 C mol^-1  × 0.03 V

= - 2894.61 J mol^-1 

= - 2.89 kJ mol^-1

Again, 

\(\triangle_tG^\ominus\)  = - 2.303 RT in K

\(\Rightarrow\) log K = - \(\frac{\triangle_tG^\ominus}{2.303RT}\)

= \(\frac{-2894.61}{2.303\times 8.314 \times 298}\)

= 0.5073

∴ K = antilog (0.5073)

= 3.2 (approximately)