Question

1. Generate a reactivity series for the metals tested. E° = 0.34 V for the copper half-cell. Use this value to calculate Eº for each of the other half reactions. Write each half reaction as a reduction and arrange them with the most powerful reducing agent at the top of the table. When arranged in this manner, a metal on the right side of an equation will spontaneously react with a species on the left hand side below it in the table. Table 1: A Displacement Series Reduction Potential (V) Half-Reaction Zn2+caq) + 2e → Z) Show all X 808 PM 7 of ENG 2020-03-28 emistry L...pdf

answer question 1 using table 1 provided bellow:

Answer 1

Answer: We have generated reactivity series for the metals tested. E0 = 0.34 V for the copper half cell. By using this value we calculate E0 for each of the other half reactions. We have written each half reaction as a reduction and have also arranged them with the most powerful reducing agent at the top of the table. When arranged in this manner, a metal on the right side of an equation will spontaneously react with species on the left hand side below it in the Table.

Now let us calculate for each metal and its pair with Cu2+

E0 = 0.34 V for the copper half cell. By using this value we calculate E0 for each of the other half reactions.

For copper  half cell we calculate from Table 1 (we calculate E0 with different element)

For Ag/Ag+ and Cu/Cu2+ pair E0 cell = E0 reduced - E0 oxidized = x - (+0.34 V) = 0.46 V

So now from Table 1 we see then we can found that for this pair value of cell potential is 459 So when we divide this value of 459 by 1000 we get E0 cell = 0.46 V

Now solving for x we have half cell potential for Ag/Ag+ = 0.46 V+0.34 V= 0.8 V(half cell potential for Ag/Ag+).

Similarly

For Fe2+/Fe3+ and Cu/Cu2+ pair E0 cell = E0 reduced - E0 oxidized = x - (+0.34 V) = 0.43 V

So now from Table 1 we see then we can found that for this pair value of cell potential is 430 So when we divide this value of 430 by 1000 we get E0 cell = 0.43 V

Now solving for x we have half cell potential for Fe2+/Fe3+ = 0.43 +0.34 = 0.77 V (half cell potential for Fe2+/Fe3+).

For I-/I2 and Cu/Cu2+ pair E0 cell = E0 reduced - E0 oxidized = x - (+0.34 V) = 0.195 V

So now from Table 1 we see then we can found that for this pair value of cell potential is 195 So when we divide this value of 195 by 1000 we get E0 cell = 0.195 V

Now solving for x we have half cell potential for I-/I2 = 0.195 V +0.34 V = 0.535 V (half cell potential for I-/I2).

For Cd/Cd2+ and Cu/Cu2+ pair E0 cell = E0 reduced - E0 oxidized = 0.34 - (x) = 0.74 V

So now from Table 1 we see then we can found that for this pair value of cell potential is 740. So when we divide this value of 740 by 1000 we get E0 cell = 0.74 V

Now solving for x we have half cell potential for Cd/Cd2+ = 0.34 V - 0.74 V = - 0.40 V (half cell potential for Cd/Cd2+).

For Pb/Pb2+ and Cu/Cu2+ pair E0 cell = E0 reduced - E0 oxidized = 0.34 - (x) = 0.47 V

So now from Table 1 we see then we can found that for this pair value of cell potential is 470. So when we divide this value of 470 by 1000 we get E0 cell = 0.47 V

Now solving for x we have half cell potential for Pb/Pb2+ = 0.34 V - 0.47 V = - 0.13 (half cell potential for Pb/Pb2+).

For Zn/Zn2+and Cu/Cu2+ pair E0 cell = E0 reduced - E0 oxidized = 0.34 V - (x) = 1.1 V

So now from Table 1 we see then we can found that for this pair value of cell potential is 1100. So when we divide this value of 1100 by 1000 we get E0 cell = 1.10 V

Now solving for x we have half cell potential for Zn/Zn2+ = 0.34 V - 1.1 V = - 0.76 V(half cell potential for Zn/Zn2+).

For Sn/Sn2+and Cu/Cu2+ pair E0 cell = E0 reduced - E0 oxidized = 0.34 V - (x) = 0.5 V

So now from Table 1 we see then we can found that for this pair value of cell potential is 500. So when we divide this value of 500 by 1000 we get E0 cell = 0.50 V

Now solving for x we have half cell potential for Sn/Sn2+ = 0.34 V - 0.5 V = - 0.16 V (half cell potential for Sn/Sn2+).

So now finally we have solve for E0 for each of the other half reactions (shown as above).

Now we are going to write each half reaction as reduction reaction as follows:

For Ag/Ag+ : Ag+ (aq) + e- $\rightarrow$ Ag (s)

For Fe2+/Fe3+ : Fe3+ (aq) + e- $\rightarrow$ Fe2+(aq)

For I-/I2   : I2 (s) + 2e- $\rightarrow$ 2 I- (s)

For Cd/Cd2+   : Cd2+ (aq) + 2e- $\rightarrow$ Cd (s)

For Pb/Pb2+ : Pb2+ (aq) + 2e- $\rightarrow$ Pb (s)

For Zn/Zn2+   : Zn2+ (aq) + 2e- $\rightarrow$ Zn (s)

For Sn/Sn2+ : Sn2+ (aq) + 2e- $\rightarrow$ Sn (s)

For Cu/Cu2+   : Cu2+ (aq) + 2e- $\rightarrow$ Cu (s)

Now we have arranged the most powerful reducing on the top of the table to the bottom of the table (as least reducing agent) as follows:

(1) Zn2+ (aq) + 2e- $\rightarrow$ Zn (s) (-0.76 V)

(2) Cd2+ (aq) + 2e- $\rightarrow$ Cd (s) (-0.40 V)

(3) Sn2+ (aq) + 2e- $\rightarrow$ Sn (s) (- 0.16 V)

(4) Pb2+ (aq) + 2e- $\rightarrow$ Pb (s) (- 0.13 V)

(5) Cu2+ (aq) + 2e- $\rightarrow$ Cu (s) (+ 0.34 V)

(6) I2 (s) + 2e- $\rightarrow$ 2 I- (s) (+0.53 V)

(7) Fe3+ (aq) + e- $\rightarrow$ Fe2+(aq) (+ 0.77 V)

(8) Ag+ (aq) + e- $\rightarrow$ Ag (s) (+0.80 V)