Question

A galvanic cell is setup with the following two half-cells: A 250.0 mL solution of 0.15 M aluminum nitrate with a platinum electrode. A 250.0 mL solution of 0.85 M cobalt(II) nitrate with a cobalt electrode. a) Determine the cell potential that will be initially measured for this galvanic cell b) How much mass is gained/lost by the cobalt electrode after the cell reaches equilibrium? c) If it takes this galvanic cell 2.00 hours to reach equilibrium, how much current would this cell generate (assume a constant current during this time)?

Please explain what you do and why.

Answer 1

Answer Question a: Potential of cell

First step: Identify the semi-reactions of oxidation and reduction and the balanced complete equation of cell

• Write the semi-reactions of reduction for Al³⁺ (from Al(NO₃)₃) to Al and for Co²⁺ (from Co(NO₃)₂) to Co. And compare their standard potentials of reduction

$Al^{3+}\, (aq)+3e^{-}\rightarrow Al\, (s)\; \; \; \; \; \; \; E^{\circ}_{red}=+1.66\, V$

$Co^{2+}\, (aq)+2e^{-}\rightarrow Co\, (s)\; \; \; \; \; \; \; E^{\circ}_{red}=+0.28\, V$

• Al³⁺ has the higher Eº_red, then it will be reduced and the the Co will be oxidized. Let's rewrite the semi-reactions into one of reduction and the other of oxidation.

$Al^{3+}\, (aq)+3e^{-}\rightarrow Al\, (s)\; \; \; \; \; \; \; E^{\circ}_{red}=+1.66\, V$

$Co\, (s)\rightarrow Co^{2+}\, (aq)+2e^{-}\; \; \; \; \; \; \; E^{\circ}_{oxid}=-0.28\, V$

For the Co, to write the semi-reaction of oxidation you just write the inverse reaction, and change the sign of Eº_red to obtain the Eº_oxid value.

• Now, it is necessary balance the chemical equation:

$\left\{\begin{matrix} \left[Al^{3+}\, (aq)+3e^{-}\rightarrow Al\, (s) \right ]\times 2 \\ \left [Co\, (s)\rightarrow Co^{2+}\, (aq)+2e^{-} \right ]\times 3 \end{matrix}\right.$

$\left\{\begin{matrix} \left2\, Al^{3+}\, (aq)+\mathbf{6\, e^{-}}\rightarrow 2\, Al\, (s) \\ 3\, Co\, (s)\rightarrow 3\, Co^{2+}\, (aq)+\mathbf{6\, e^{-}} \end{matrix}\right.$

$\mathbf{2\, Al^{3+}\, (aq)+3\, Co\, (s)\rightarrow 2\, Al\, (s) +3\, Co^{2+}\, (aq)}$

Second step: Determinate the Eº of the galvanic cell

• The standard potential of cell is defined as:

$E^{\circ}=E^{\circ}_{red}+E^{\circ}_{oxid}$

$E^{\circ}=+1.66\, V+\left (0.28\, V \right )=\boldsymbol{\mathbf{+1.38\, V}}$

*This is in standard conditions: 298 K and solutions 1M

Third step: Determinate the E of the galvanic cell if you use 250 mL of Al(NO₃)₃ 0.15 M and 250 mL of Co(NO₃)₂ 0.85 M

1. At this conditions, we must use the Nersnt equation to determinate the potential of cell

$E=E^{\circ}-\frac{RT}{nF}ln\left ( Q \right )=E^{\circ}-\frac{RT}{nF}ln\left ( \frac{\left [ Co^{2+} \right ]^{3}}{\left [ Al^{3+} \right ]^{2}} \right )$

where

• R = gas constant = 8.314 J/mol.K
• T = absolute temperature = 298 K
• n = moles of transfered electrons = 6 (see the balance of semi-reactions)
• F = faraday constant = 96485 J/V.mol
• Q = reaction quotient (only consider gas and aqueous species)

2. Then, we need to find the [Co²⁺] and [Al³⁺] in the cell:
   • The Co²⁺ comes from 250 mL of Co(NO₃)₂ 0.85 M

$Co(NO_{3})_{2}\rightarrow Co^{2+}+2\, NO_{3}^{-}$

1 mol of Co(NO₃)₂ produces 1 mol of Co²⁺, then, [Co(NO₃)₂] = [Co²⁺] = 0.85 M

• The Al³⁺ comes from 250 mL of Al(NO₃)₃ 0.15 M

$Al(NO_{3})_{3}\rightarrow Al^{3+}+3\, NO_{3}^{-}$

1 mol of Al(NO₃)₃ produces 1 mol of Al³⁺, then, [Al(NO₃)₃] = [Al³⁺] = 0.15 M

3. Now, we can calculate the potential of cell:

$E=E^{\circ}-\frac{RT}{nF}ln\left ( \frac{\left [ Co^{2+} \right ]^{3}}{\left [ Al^{3+} \right ]^{2}} \right )=+1.38\, V-\frac{8.314\, J/mol.K\times 298K}{6\times 96485\, J/C.mol}ln\left ( \frac{\left (0.85 \right )^{3}}{\left (0.15 \right )^{2}} \right )$

$E=+1.38\, V-4.28x10^{-3}\, V\times ln\left ( 27.294 \right )=+1.38\, V-4.28x10^{-3}\, V\times3.307$

$E=+1.38\, V-0.014\, V=\mathbf{+1.37\, V}$

Answer Question b: mass gained/lost by cobalt electrode

First step: Calculate the equilibrium constant of the reaction:

• At equilibrium the standard Gibbs free energy is zero, and the free energy is defined as:

$\Delta G^{\circ}=-nFE^{\circ}=-RT\, ln\left ( K \right )=0$

then

$-nFE^{\circ}=-RT\, ln\left ( K \right )\; \; \; \Rightarrow \; \; \; ln\left ( K \right )=\frac{nFE^{\circ}}{RT}$

$K=e^{\frac{nFE^{\circ}}{RT}}=e^{\frac{6\times 96485\, J/C.mol\times 1.38\, V}{8.314\, J/mol.K\times 298K}}$

$K=e^{322.45}=\mathbf{1.09x10^{140}}$

Second step: Determinate the reaction direction

• From question a, we have calculate the reaction quotient for the cell with 250 mL of Al(NO₃)₃ 0.15 M and 250 mL of Co(NO₃)₂ 0.85 M

$Q = \frac{\left [ Co^{2+} \right ]^{3}}{\left [ Al^{3+} \right ]^{2}} =\frac{\left (0.85 \right )^{3}}{\left (0.15 \right )^{2}}=27.294$

• In this case, Q << K (27.294 << 1.09x10¹⁴⁰). This means, that reaction must go to right to reach the equilibrium.
• According to the reaction, if reaction goes to right, the Co electrode will be consumed ang it will loss mass

Third step: Determinate the amount of Co²⁺ produced

• All mass lost by Co electrode will be transformed in Co²⁺. Then, we need determinate the [Co²⁺] at equilibrium and compare with the initial [Co²⁺]. to do this, we made a ICE table:

	2 Al3+	+	3 Co	-->	2 Al	+	3 Co2+
Initial	0.15		--		--		0.85
Change	-2x		--		--		+3x
Equilibrium	0.15-2x		--		--		0.85+3x

• The equilibrium constant for this reaction is:

$K = \frac{\left [ Co^{2&plus;} \right ]^{3}}{\left [ Al^{3&plus;} \right ]^{2}}= \frac{\left (0.85&plus;3x \right )^{3}}{\left (0.15-2x \right )^{2}}= 1.09x10^{140}$

• Like thw equilibrium constant is very large, we can made an approximation. A very large K indicates that almost reactants will be convert to products, then,

$\left [ Al^{3&plus;} \right ]_{initial}\approx \left [ Al^{3&plus;} \right ]_{change}$

$0.15\approx 2x\; \; \; \; \; \; \Rightarrow \; \; \; \; \; \; x\approx \frac{0.15}{2}=\mathbf{0.075\, M}$

• Now, we can calculate the amount of Co²⁺ produced, it will be equal to the change in [Co²⁺] = 3x

$\left [ Co^{2&plus;} \right ]_{produced}=3x=3\left ( 0.075\, M \right )=\mathbf{0.225\, M}$

the moles produced of Co²⁺ are (the cell has 250 mL = 0.25 L):

$\left [ Co^{2&plus;} \right ]=\frac{n_{Co^{2&plus;}}}{V}\; \; \; \; \; \Rightarrow \; \; \; \; \; n_{Co^{2&plus;}}=\left [ Co^{2&plus;} \right ]\times V$

$n_{Co^{2&plus;}}=0.225\, mol/L\times 0.25\, L=\mathbf{0.0563\, mol}$

Fourth step: mass of Co lost

• All moles of Co²⁺ produced comes from Co electrode:

${2\, Al^{3&plus;}\, (aq)&plus;3\, Co\, (s)\rightarrow 2\, Al\, (s) &plus;3\, Co^{2&plus;}\, (aq)$

• According to reaction, 1 mol of Co produces 1 mol of Co²⁺. Then if 0.0563 mol of Co²⁺ were produced, this means that 0.0563 mol of Co were consumed
• Therefore, the mass lost by Co electrode is

$n_{Co}=\frac{mass_{Co}}{AW_{Co}}\; \; \; \; \; \Rightarrow \; \; \; \; \; mass_{Co}=n_{Co}\times AW_{Co}$

$mass_{Co}=0.0563\, mol\times 58.933\, g/mol=\mathbf{3.315\, g}$

Answer Question c: current generated

• Data
  • 2 hours to reach equilibrium
  • 3.315 g of Co lost (0.0563 mol of Co)

• We can use the Faraday's laws to determinate the amount of current generated:

$m=\frac{I\times t\times MW}{F\times z}\; \; \; \; \; \Rightarrow \; \; \; \; \; n=\frac{m}{MW}=\frac{I\times t }{F\times z}$

where

• m = mass lost by Co electrode = 3.315 g
• MW = molar mass of Co = 58.933 g/mol
• n = moles of Co = 0.0563 mol
• I = current generated (Amp)
• t = time to reach the equilibrium (seg) = 2 h = 7200 seg (2 h x 3600 seg/h)
• F = Faradys constant = 96485 C/mol
• z = valence number of Co²⁺ = 2

then, the current generated is

$n=\frac{I\times t }{F\times z}\; \; \; \; \; \Rightarrow \; \; \; \; \; I=\frac{nFz}{t}$

$I=\frac{0.0563\, mol\times 96485\,C/mol\times 2}{7200\, seg}=\mathbf{1.509\, amp}$

RESUME

• Question a : Potential of cell = +1.37 V
• Question b : the mass lost by Co electrode after the cell reaches the equilibrium is 3.315 g
• Question c: the cell will generate 1.509 amp