# EMF and electron flow

1. Apr 29, 2015

### Raghav Gupta

1. The problem statement, all variables and given/known data
A variable , opposite external potential $E_{ext}$ is applied to the cell Zn | Zn2+ (1M) || Cu2+ (1M) | Cu, of potential 1.1V . When Eext < 1.1 V and Eext > 1.1V , respectively electrons flow from :
1. anode to cathode and cathode to anode
2. cathode to anode and anode to cathode
3. cathode to anode in both cases
4. anode to cathode in both cases

2. Relevant equations

NA
3. The attempt at a solution
At anode oxidation happens and at cathode reduction.
So electrons are moving from anode to cathode when Eext < 1.1 V and cathode to anode when Eext > 1.1 V
So option 1 is looking correct.
In answer key option 4 that is anode to cathode in both cases is given.
Why?

2. Apr 29, 2015

### Merlin3189

My guess is this is almost a "trick" question,
The anode is defined as the electrode where electrons leave a cell and the cathode where electrons enter a cell. So no matter which way the current flows, external to the cell electrons always flow from anode to cathode.
When you switch the battery, you switch the names of the electrodes.

3. Apr 29, 2015

### Raghav Gupta

Thanks, got the tricky part.

4. May 13, 2015

### James Pelezo

Look at a table of Reduction Potentials (here's one for quick reference: http://chemunlimited.com/Table of Reduction Potentials.pdf ), in all cases electrons flow from the more negative reduction potential to the more positive reduction potential ... ALWAYS. Choose any two half reactions and note the Eo-values. The more negative value is the anode (site of oxidation) and the more positive value cathode (site of reduction). I might suggest studying how the respective ion concentrations in each half cell affect the voltage of the Galvanic/Voltaic Process. That is, what voltage does the cell have if [reducing agent ion] < [oxidizing agent ions] and visa versa. If [oxidizing agent ions] = [reducing agent ions] what is the cell voltage? Happy charging...

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