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Untitled - Kelly Walsh High School

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Electrochemistry 271<br />

We can use this table of standard reduction potentials to write the overall cell<br />

reaction and to calculate the standard cell potential, the potential (voltage)<br />

associated with the cell at standard conditions. There are a couple of things to<br />

remember when using these standard reduction potentials to generate the cell<br />

reaction and cell potential:<br />

1. Since the standard cell potential is for a galvanic cell, it must be a positive<br />

value, E > 0.<br />

2. Since one half-reaction must involve oxidation, we must reverse one of the<br />

half-reactions shown in the table of reduction potentials in order to indicate<br />

the oxidation. If we reverse the half-reaction, we must also reverse the sign<br />

of the standard reduction potential.<br />

3. Since oxidation occurs at the anode and reduction at the cathode, the standard<br />

cell potential can be calculated from the standard reduction potentials of<br />

the two half-reactions involved in the overall reaction by using the equation:<br />

E cell E cathode – E anode > 0<br />

But remember both the E cathode and E anode values are shown as reduction<br />

potentials, used directly from the table without reversing.<br />

Once you have calculated the standard cell potential, then the reaction can be<br />

written by reversing the half-reaction associated with the anode (show it as oxidation)<br />

and adding the two half-reactions.<br />

Don’t forget that the number of electrons lost must equal the number of electrons<br />

gained. If they are not equal, use appropriate multipliers to ensure that<br />

they are equal.<br />

Calculate the potential of a cell using the following half-cells:<br />

Ni2 2 e l Ni(s) E 0.25 V<br />

Ag e l Ag(s) E 0.80 V<br />

First, calculate the cell potential using:<br />

E cell E cathode E anode > 0<br />

Since the cell potential must be positive (a galvanic cell) there is only one<br />

arrangement of 0.25 and 0.80 V than can result in a positive value:<br />

Ecell 0.80 V (0.25 V) 1.05 V<br />

This means that the Ni electrode is the anode and is involved in oxidation.<br />

Therefore, we reverse the reduction half-reaction involving Ni, changing the

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