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Determine the cell potential for an electrochemical cell based on the following two half-reactions (assume T = 298K): Oxidation: Cu(s) + Culata . + 2 e E° -0.34 V (aq, 0.060 M) Reduction: Mno 4(aq, 1.636 M) + 4H+ (aq, 2.000 M) + 3e + MnO2(s) + 2 H2O(1) Eº = 1.68 V Ecell number (rtol=0.03, atol=1e-08)

Sagot :

The cell potential for the reaction is 28V.

The first step is to ascertain the cell potential at the standard conditions of 1 mol/L concentrations and 1 atm pressures at 25°C.

It goes like this:

The cell's oxidation and reduction half-reactions should be written down.

To find the reduction potential, E0red for the reduction half-reaction, look it up in a table of reduction potentials.

The potential difference between a cell's two electrodes, known as the cell potential, results from the transfer of electrons through a cell's external circuit when the cell has not yet reached equilibrium.

The voltage that exists between a battery's two half cells can be determined using the cell potential.

To find the oxidation potential, reverse the sign and look up the reduction potential for the opposite of the oxidation half-reaction.

Cell potential =Eo/E

E=2xRt=2x0.03=0.06

Cell potential =1.68/0.06=28V

To learn more about cell potential-

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