A.) What is the value of E^o of the cell? (o refers to standard) B.) What is the

Question

A.) What is the value of E^o of the cell? (o refers to standard)

B.) What is the oxidation state of Cr in Cr2O7 ^-2?

C.) What is the value of the reaction quotient, Q? Make sure net reaction is properly balanced

D.) How many electrons are transfered in the balanced reaction? (i.e., what is the value of n in the Nernst equation?)

E.) What is the value of the cell potential, E, at 25 degrees C?

motor 0.1 M MnO4 0.2 M Mn2+ 0.01 M H 0.4 M Cr3+ 0.3 M Cr202 0.01 MH Salt Bridge The standard reduction potentials are as follows MnO4-+ 8H+ + 5e- Mn2+ + 4H20 CnO,2 + 14H+ + 6e- 2CP+ + 7H20 E"= 1.51 V Eo1.33 V

Explanation / Answer

The cell potential is defined as

E cell = E reduction + E oxidation

The MnO4 has a stronger reduction potential than the other reaction so

reduction reaction will be

MnO4- + 8H + 5e- === Mn+2 + 4H2O E = 1.51V

2Cr+3 + 7 H2O ==== Cr2O7-2 + 14H+ + 6e- E = -1.33V

so E cell = 1.51 + 1.33 = 0.18 V

b) oxidation state of Cr2O7-2 , to solve this remember that oxygen has -2 so the sum of all oxidation states must be equal to -2

2x - 14 = -2

2x = 14 - 2 = 12

x = 6, the oxidation state of Cr is +6

c) balance the equation, to do this lets multiply each equation by the number of electrons of the other equation

(MnO4- + 8H + 5e- === Mn+2 + 4H2O) * 6

(2Cr+3 + 7 H2O ==== Cr2O7-2 + 14H+ + 6e- ) * 5

---------------------------------------------------------------------

6 MnO4- + 48H + 30e- === 6Mn+2 + 24 H2O

10 Cr+3 + 35 H2O ==== 5 Cr2O7-2 + 70 H+ + 30 e-

------------------------------------------------------------------------

10 Cr+3 + 6 MnO4- + 35 H2O + 48H + 30e- ==== 5 Cr2O7-2 + 70 H+ + 30 e- + 6Mn+2 + 24 H2O

10 Cr+3 + 6 MnO4- + 11 H2O ===== 5 Cr2O7-2 + 22 H+ + 6Mn+2

For a reaction like

aA + bB cC + dD

Equilibrium constant is the relationship of products and reactants expressed like this

K = [A]a [B]b / ( [C]c[D]d ), the reaction quotient is very similar just put the values from the statement

Q = ( [Cr2O7]5[H]22 [ Mn]6  ) /  [Cr]10 [MnO4]6

Q = (0.35 * 0.0122 * 0.26) / (0.410 * 0.16)

i am using a spreadsheet to solve this I am getting a value of 1.483 x 10-41

d) 30 electrons are being transferred in the reaction

E) E = Eo cell - 0.0592 / n * log Q, n is electrons transferrred, this equation is only valid at 25C

E = 0.18 - (0.0592 / 30) * log (1.483 x 10-41)

E = 0.26 V

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