For the reaction given below the [H + ] is increased from 1.00 M to 3.00 M. What

Question

For the reaction given below the [H+] is increased from 1.00 M to 3.00 M. What happens to the voltage? Write i for increases, n for no change, and d for decreases.

a)For the above reaction the [Zn2+] is decreased from 1.00 M to 0.30 M. What happens to the voltage?

b)For the above reaction the H2 pressure is increased from 1.00 atm to 4.2 atm. What happens to the voltage?

c)For the above reaction the Cl- concentration is increased from 1.00 M to 4.50 M by the addition of NaCl. What happens to the voltage?

                 d)For the Zn-H2 cell, what is the voltage if [Zn2+] = 2.72 M, [H+] = 0.0170 M, and the hydrogen pressure is 157.7 atm

For the reaction given below the [H+] is increased from 1.00 M to 3.00 M. What happens to the voltage? Write i for increases, n for no change, and d for decreases. gavalnic cell LeChatlier For the above reaction the [Zn2+] is decreased from 1.00 M to 0.30 M. What happens to the voltage? For the above reaction the H2 pressure is increased from 1.00 atm to 4.2 atm. What happens to the voltage? For the above reaction the Cl- concentration is increased from 1.00 M to 4.50 M by the addition of NaCl. What happens to the voltage? For the Zn-H2 cell, what is the voltage if [Zn2+] = 2.72 M, [H+] = 0.0170 M, and the hydrogen pressure is 157.7

Explanation / Answer

Zn(s) + 2H+(aq) -------> Zn2+(aq) + H2(g)

Ecell = E0 cell - 0.0591/n log([Zn2+]*P(H2)/[H+]^2)

Increasing [H+] from 1M to 3M Voltage will increase

(a) Decrease in [Zn2+] from 1M to 0.3M the voltage will increase

(b) Voltage will decrease if P(H2) is increased

(c) It does not depend on [Cl-] hence No Change

(d)Ecell = 0.76 - 0.0591/2 log(2.72*157.7/(0.017)^2)

Ecell = 0.5776 V

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