Explore BrainMass
Share

# Electrochemistry

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

1. Calculate the standard stat emf, fancy E, change in G, and K at 25 degrees celcus for the following reaction

2Na(s) + 2H20(l) --------- 2NaOH(aq) + H2(g)

standard reduction potentals:
Na(positive 1) + e- ---------Na -2.71V
2H20 + 2e- ------------H2 + 2OH- - .83V

2. What mass of Pb would be deposited by electrolysis of and aqueous solution of lead(II) nitrate using a current of 100 amps of 8.00 hours?

3. Draw a complete galvanic cell in which the electrodes are a nickel bar dipping into a 0.100M solution of Ni(positive 2) and a manganese bar dipping inot a 0.500 M soulution of Mn(positive 2). Label all the parts of the cell and indicate the direction of electron flow. Write a balanced eqation of the cell reaction and calculate the emf and fancy E for the cell

Standare reduction potentials:

Ni/Ni(positive 2) = -0.25 V Mn/Mn(positive 2) = -1.18V

4. Balance the following redox equations

Cr2O7(negative 2)(aq) + I2(aq)--------- Cr (positive 3) + IO3(negative 1)(aq)
(acidic solution)

HS2O3(negative 1)(aq)---------- S(s) + HSO4(negative 1) (aq) (acidic solution)

© BrainMass Inc. brainmass.com October 9, 2019, 5:33 pm ad1c9bdddf
https://brainmass.com/chemistry/electrochemical-cells/electrochemistry-standard-emf-60951

#### Solution Preview

Please see the attached file.

1. 2Na(s) + 2H20(l)  2NaOH(aq) + H2(g)

Standard reduction potentials:
Na+ + e-  Na; -2.71V
2H20 + 2e-  H2 + 2OH-; -0.83V

E = 2.71*2 - 0.83 = 4.59 V

Free energy ΔGº = -nFEºcell, wheren is the number of moles of electrons transferred
F is Faraday's constant and represents the charge of one mole of electrons = 96,500 coulombs/mol
Here, ΔGº = -2*96,500*4.59 = 885.87 KJ
Equilibrium Constant: Kc = enFEº/RT = exp(885870/(8.31*298)) = exp(31767661)

2. What mass of Pb would be deposited by electrolysis of and aqueous solution of lead (II) nitrate using a current of 100 amps of 8.00 hours?

The reduction of Pb2+ ions will occur according to the following equation:

Pb2+(aq) + 2e-  ...

#### Solution Summary

It provides examples of solving electrochemistry questions.

\$2.19