Question

3. The literature reduction potential under standard conditions for each metal used in this experiment was obtained from standard reduction tables. The literature reduction % cell potentials are in order of increasing reduction potential (least able to be reduced to most able to be reduced). Use the information in question 1 and 2 to list the metals tested in order of increasing reduction potential. Place this information in the table below. Literature Reduction Potentials Literature Voltage Your list of increasing reduction potential. A 3 e > Al(s) C2aa) + 2e-Cu( -1.66 V SnCs) Zn2 2 e Zn (s) Sn2t caa) +2e -0.76 V F caa) 2-re Fe2 +2 e > Fe (s) -0.44 V A1(aa) 2 AIC) -0.14 V Sn2+ 2 e> Sn (s) 2n CS) +0.34 V 2n2 (aa) + 2 Cu2 2 e> Cu (s) 4. Compare the list from the literature to your list. Are the two lists the same or different? Explain any experimental variables that may have caused differences. 5. What voltage was obtained for the Znel Zn21.OMl Cu(1.O) Cus) Cell? V What voltage was obtained for the Znal Zn20.10M)|| Cu'0 10M) Cu) cell? +0. 50 Would you expect to have the same voltage or different voltages for these two cells? Explain your answer.

Answer 1

3. Answer three is you already written, it is based on your question 1 and 2. The reason of increasing is as the standard reduction potential move toward higher value number (toward positive counting), the reduction ability is increasing, so literature order is this.

4. The list is same, because increasing is as the standard reduction potential move toward higher value number (toward positive counting), the reduction ability is increasing, but in your list you do wrong. there is no experimental variations and literature.
5. The cell have same voltage 1.1 V , the explanation is given below

E_cell = E_o^cell - 2.303(RT/nF)*log(Product/Reactant)

Where E_cell is cell potential

E_o^cell is standard cell potential

n is the number of electron take part in the reaction.

F = faraday constant, the charge of one mole electron= 96487 Cmol^-1

Product and Reactant concentration.

The value of 2.303RT/F = 0.059

Note: I am taking room temperature condition 298K.

The value 0.059 putting in above equation we get

E_cell = E_o^cell - 0.059/n*log(Product/Reactant)

The cell net reaction is

Zn + Cu²⁺ → Zn²⁺ + Cu

n= 2

E_o^cell = E_oCu²⁺ - E_oZn²⁺ = 0.34-(-0.76) = 1.1V

Now for first reaction

E_cell = 1.1 - 0.059/2*log(Zn²⁺* Cu)/(Zn*Cu²⁺)       ...............................................................(1)

Cu and Zn = 1 because activity or concentration is 1 and also concentration of copper 2+ and Zinc 2+ ion is 1.0M given

putting in above

E_cell = 1.1 - 0.059/2*log(1/1)= 1.1 V

Note: log1 = 0

Now for second reaction

Cu and Zn = 1 because activity or concentration is 1 and concentration of copper 2+ and Zinc 2+ ion is 0.1M given

putting in equation number one

E_cell = 1.1 - 0.059/2*log(0.1M)/(0.1M) = 1.1 V

Note: log1 = 0

Hence both have same value 1.1 V .