Electrochem For the electrochemical cell with the following redox reaction, Ni(s
ID: 1000300 • Letter: E
Question
Electrochem
For the electrochemical cell with the following redox reaction, Ni(s)+2H+(aq, 1M)- Ni^2+(aq, 1 M) + H_2(g, 1 atm), which one of the following changes will cause a decrease in the cell potential, E_cell? Increase the [H^+] in the cathode (to 1.5 M for example). Decrease the mass of the nickel electrode. Decrease the concentration of Ni^2+ ions (to 0.5 M for example). Increase the pressure of H_2 from 1.0 atm to 2.0 atm. None of the above. Which, if any, of the following metals would not be capable of acting as a sacrificial anode tor iron, Fe? The standard reduction potential for Fe is E degree = -0.44 V and the standard reduction potentials for the M^2+/M half reactions are given below. Cadmium, Cd, E degree = -0.40 V Magnesium, Mg, E degree = -2.37 V Manganese, Mn, E degree = -1.18 V Zinc, Zn,E degree =-0.76 V All of these are capable of acting as sacrificial anods with ironExplanation / Answer
34) For the reaction, Ecell = Eocell -log (Q)*0.0592/n
Where Q = [Ni2+]/[H+]2
In option A, by increasing the concentration of H+, Q will decrease, and thus, Ecell will increase. So, this option is wrong.
In option C, by decreasing the concentration of Ni2+, Q will decrease, and thus, Ecell will increase. So, this option is also wrong.
By changing the mass of Ni and pressure of H2 will not affect the Q, because they are not present in the Q equation. So, option B and D are also not correct.
Answer is E, i.e. none of above.
35) Standard potential of Fe is -0.44 V.
Fe is acting as cathode in reaction. If the anode is acting as sacrificial, that means it is being oxidised in the reaction. To be oxidised, the reduction potential of the anode should be more negative. All of the metals given in the options have more negative reduxtion potential. So, answer is E.
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