Question

Calculate the value of \[\ce{E^\circ}\]cell, E cell and ΔG that can be obtained from the following cell at 298 K.

\[\ce{Al/Al^3+ _{(0.01 M)} // Sn^{2+} _{(0.015 M)}/Sn}\]

Given: \[\ce{E^\circ  Al^3+/Al = -1.66 V; E^\circ\phantom{.}Sn^2+/Sn = -0.14 V}\]

Answer 1

\[\ce{Al/Al^3+ _{(0.01 M)} // Sn^{2+} _{(0.015 M)}/Sn}\]

Given, \[\ce{E^\circ  Al^3+/Al = -1.66 V}\]

\[\ce{E^\circ\phantom{.}Sn^2+/Sn = -0.14 V}\]

The value of \[\ce{E^\circ}\] is calculated by the formula,

\[\ce{E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}}\]

or

\[\ce{E^\circ_{cell} = E^\circ_{oxidation} - E^\circ_{reduction}}\]   ...\[\ce{[2Al + 3Sn^2+-> 2Al^3+ + 3Sn]}\]

\[\ce{E^\circ_{cell}}\] = − 0.14 − (− 1.66)

\[\ce{E^\circ_{cell}}\] = − 0.14 + 1.66

∴ \[\ce{E^\circ_{cell}}\] = 1.52  V

Cell reaction is: 

\[\ce{2Al + 3Sn^2+ → 2AI^3+ + 3Sn}\] and n = 6

According to Nernst equation, at 298 K 

\[\ce{E_{cell} = E°_{cell} - \frac{0.059}{n} log  \frac{([Al^{3+}]^2)}{([Sn^{2+}]^3)}}\]

`=1.52 - 0.059/6 log  ([0.01]^2)/([0.015]^3)`

= 1.52 - 0.01447

∴ `E_(cell) = 1.50553` V.

`Delta G = -"nF""E"_"cell"^\circ`

Where,

n = 3 moles of electrons

F = 96485 C/mol

Convert volts to joules per mole:

1 V = 1 J/C

Now, Calculate ΔG:

ΔG = −(3 × 96485 C/mol) × (1.52 J/C)

ΔG = −(3 × 96485 × 1.52) J/mol

ΔG = − 440079.8 J/mol

∴ ΔG = − 871742 Joules