1 decrease potential 2 no change in potential 3 decrease potential 2 no change i

Question

1

decrease potential

2

no change in potential

3

decrease potential

2

no change in potential

3

increase potential Consider a galvanic cell based on the following line notation at 277 K: where the standard cell potential is 0.71 V. What will the following changes do to the potential of the cell? decrease poteniaRemove water from the anode portion of the cell (concentrating the solution) no change in potentialEqually concentrate both the anode and cathode portions of the cell decrease potential Increase the mass of the Al Add hint text here. Submit Answer Incorrect. Tries 2/3 Previous Tries

Explanation / Answer

Mg/Mg+2//Al+3/Al

As Magnesium standard reduction potential is less compared to aluminium so Magnesium acts as reducing agent and undergo oxidation at anode

At anode : Oxidation : Loss of electrons. Magnesium acts as reducing agent

Mg(s) ---------> Mg+2(aq)+ 2e- +2.37 V

At cathode : Reduction : Gain of electrons. Aluminium acts as oxidising agent

Al+3(aq) + 3e- --------> Al(s) -1.66 V

Net ionic equation : 2 (  Al+3(aq) + 3e- --------> Al(s) ) ===> 2Al+3(aq) + 6e- ----> 2Al(s)

3 ( Mg(s) ---------> Mg+2(aq)+ 2e- ) ===> 3Mg(s) -------> 3Mg+2(aq) + 6e-

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2Al+3(aq)   +  3Mg(s) --------> 2Al(s) + 3Mg+2(aq)

we know Nernst equation to find out the cell potential

Ecell = (Eo cell) - (0.0592/n) log10 Q

Ecell = (Eo cell) - (0.0592/n) log10 [oxi/red]

Ecell = (Eo cell) - (0.0592/n) log10 [Mg+2]3/[Al+3]2

In the above reaction n is number of electrons = 6

Q is reaction quotient = products/reactants

1) When water is removed from anode portion of the cell this leads to concentrating the magnesium solution i.e increases the concentration of Magnesium solution at anode decreases the cell potential because this leads to the increase in the Q value (greater than 1) that decreases cell potential

For better understanding consider Nernst equation :

For example let us consider the concentration of Magnesium is increased 10 times to that of aluminium then

[Mg+2] = 10 [Al+3]

   Ecell = (Eo cell) - (0.0592/n) log10 [Mg+2]3/[Al+3]2

     Ecell = 0.71 V - (0.0592/6) log10 [10)3/12]

Ecell = 0.71 V - 0.0099 X 3

  Ecell = 0.71 V - 0.0297

Ecell = 0.6803 V

As Q > 1 then Ecell decreases

2) Equally concentrating both anode and cathode : This infers that both the concentrations of Magnesium as well as aluminium are equal then the cell potential slightly decreases because this leads to the increase in the Q value (greater than 1) that decreases slightly cell potential

For better understanding consider Nernst equation :

Let us consider the concentration of Magnesium and aluminium be 10 M

[Al+3] = [Mg+2] = 10 M

then    Ecell = (Eo cell) - (0.0592/n) log10 [Mg+2]3/[Al+3]2

  Ecell = (Eo cell) - (0.0592/n) log10 [10]3/[10]2

   Ecell = 0.71 V - (0.0592/6) log10 [1000/100]

Ecell = 0.71 V - 0.0099 X 1

Ecell = 0.70001 V

As Q >1 the Ecell decreases but slightly

3) When mass of aluminium is increased then the concentration of aluminium is increased then it increases the cell potential because this leads to the reduction of Q value (less than 1) that increases cell potential

For better understanding consider Nernst equation :

Let us consider the concentration of aluminium is increased to 10 units than that of magnesium

[Al+3] = 10 [Mg+2]

then   

Ecell = (Eo cell) - (0.0592/n) log10 [Mg+2]3/[Al+3]2

  Ecell = (Eo cell) - (0.0592/n) log10 [1]3/[10]2

   Ecell = 0.71 V - (0.0592/6) log10 [1/100]

Ecell = 0.71 V - (0.0099 X -2)

Ecell = 0.71 V + 0.0198

Ecell = 0.7298 V

As Q value is less than 1 then Ecell increases

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