# Calculations with Single Voltaic Cells & Electrolytic Cells

Instructor: Dave Hays
What are the differences between voltaic and electrolytic cells? How are the equations for the reactions written? In this lesson, learn about these equations and how to write and interpret the chemical equations for the reactions.

## Problem Solving

Ann and Beth are working together to determine how to calculate the voltage of a simple voltaic cell. Ann has done these calculations before and is helping Beth set up the problem. Beth starts out with the two half-cells. ''Ok, we have two half-cells, copper metal in a copper (II) sulfate solution and zinc metal in a zinc sulfate solution.''

''Correct,'' replies Ann. ''What is the tube called that connects the two half-cells?''

After pausing for just a brief moment, Beth states, ''It's a salt bridge, which is used to maintain charge neutrality in the two half-cells.''

''Again, correct,'' Ann said.

''Now, the first thing I need to do is determine what is being oxidized and what is being reduced.''

''That's right,'' says Ann. ''Now, tell me what those two terms mean.''

''Well, in oxidation a metal loses electrons and its charge increases to form a positively charged metal ion called a cation, while in reduction a metal ion gains electrons to form a neutral metal atom. Remind me again how I work with these equations.''

Ann responds with a three-word hint, ''Standard reduction potentials.''

Beth chimes in, ''Yes, I look up the two reduction half-reactions from the table of standard reduction potentials and compare them.''

## Standard Reduction Potentials

Cu+2 (aq) + 2 e- → Cu(s)

E0 (V) = + 0.15

Zn+2 (aq) + 2 e- → Zn(s)

E0 (V) = - 0.76

The half-reactions are written as reduction reactions by convention and each reduction reaction potential is calculated relative to the reduction of hydrogen, which is set at zero volts.

## Standard Reduction Potential Explanation

Ann continues to help her friend finish the problem. ''Now, please explain to me what these equations mean.''

Beth explains from memory. ''In the case of the first equation, the copper-two cation gains two electrons to form copper metal. This is a reduction half-reaction because the charge on the copper-two ion is decreasing. The standard reduction potential of positive 0.15 volts means that the reaction will favor the formation of the copper metal at equilibrium. The second equation shows that the zinc-two cation also gains two electrons to form zinc metal. This also is a reduction half-reaction because the charge on the zinc-two ion also goes to zero. The negative standard reduction potential for this half-reaction indicates that the reaction favors the reactants at equilibrium.''

''Now, why are they called half-reactions?'' Ann inquires.

''That's because each reaction is half of a balanced chemical equation for a reaction.''

''Yes,'' states Ann, ''Next, you need to look at the two half-reactions. The one that has the more positive potential you leave as it is. The other you switch, so that it is now an oxidation half-reaction, not reduction. Remember to change the sign on the potential. Then, you add them together to get the complete equation for the reaction.''

''You mean, like this?'' says Beth.

Cu+2 (aq) + 2e- → Cu(s)

E0 (V) = + 0.15

Zn(s) → Zn+2 (aq) + 2e-

E0 (V) = 0.76

Cu+2 (aq) + 2e- + Zn(s) + → Zn+2 (aq) + 2e- + Cu(s)

E0 (V) = 0.76 + 0.15 = 0.91

''Now, do you see that you can cancel out the electrons from both sides of the equation for the reaction?''

''Yes, I do.'' responds Beth. ''So, now the equations for the reaction become,''

Cu+2 (aq) + Zn(s) → Zn+2 (aq) + Cu(s)

E0 (V) = 0.91

''Can you tell me how to interpret this equation?'' queried Ann.

''Sure,'' replied Beth. ''In this reaction, the copper-two ion is reduced by zinc metal to form the zinc-two ion and copper metal. The copper-two ion is oxidized and the zinc metal is reduced. The standard reduction potential for the reaction is 0.91 volts which means the reaction is spontaneous and the equilibrium favors the formation of the products.''

''Now, let's extend our knowledge by examining electrolytic cells.''

## Electrolytic Cell

''Remember,'' says Ann, ''in the voltaic cell the copper was the cathode because reduction occurred in that half-cell and the zinc was the anode because oxidation occurred in that half-cell. And because of the differences in the standard reduction potentials of the two meals the reaction was spontaneous in the direction indicated by the equation you wrote. Does that still make sense?''

''Now,'' Ann goes on, ''Suppose we want to run the reaction backward, reducing the zinc-two ion and oxidizing the copper metal. Could we?''

''No, we could not, because it is not spontaneous in that direction,'' is Beth's reply.

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