The diagram below represents a voltaic cell. Which one of the statements about t
ID: 1064233 • Letter: T
Question
The diagram below represents a voltaic cell. Which one of the statements about this cell is NOT correct? Zn(s) Zn^2+ (1.0 M)(Cu^2+ (1.0 M) Cu(s) The mass of the zinc electrode decreases during discharge. The copper electrode is the anode. Electrons flow through the external circuit from the zinc electrode to the copper electrode. Reduction occurs at the copper electrode during discharge. The concentration of copper ions decreases during discharge. Delta G degree = 1.8 kJ/mol at 25 degree C for a reaction. What is the equilibrium constant (K) for the reaction. (R = 8.314 J/mol middot K) An acid solution is titrated with a base. Calculate the pH after 35.0 mL of 0.200 M NaOH solution has been added to 20.0 mL sample of 0.200 M HBr solution. What is the pH of a buffer made by combining 150 mL of 0.25 M NaC_2 H_3 O_2 with 150 mL of 0.45 M HC_2 H_3 O_2? The K_a of acetic acid is 1.75 times 10^-5.Explanation / Answer
53) (a) Here the Zinc will undergo oxidation and will gets consumed and its mass will decrease [Correct]
b) copper is cathode not anode, It is undergoing reduction
c) Electrons will flow from anode to cathode [so correct]
d) Yes reduction at cathode
e) the copper ion will get reduced to copper metal and hence the concentration of copper ions will reduce
54)
We know that
Delta G0 = RTnlKeq
Given :
Delta G0 = 1.8 KJ / mol
R = gas constant = 8.314 X 10^-3 KJ / mol K
T = temperature = 298 K
Keq = ?
Putting values
1.8 = - 8.314 X 10^-3 X 298 X ln Keq
ln Keq = -0.726
Keq = antilog of (-0.726)
Keq = 0.483
55) The acid will react with base in 1:1 molar ratio as
NaOH + HBr --> NaBr + H2O
Moles of acid present = Molarity of acid X volume of acid = 0.2 X 20 millimoles = 4millimoles
Moles of base present = Molarity of base X volume of base = 0.2 x 35 = 7 millmoles
so millmoles of base neutralized = 4 millimoles
Millmoles of base left = inital moles - moles neutralzied = 7 - 4 = 3 millmoles
concentration of base = [OH-] = moles /Total volume = 3 millimoles / 55 mL = 0.055 molar
pOH = -log[OH-] = 1.26
pH = 14 -pOH = 14 - 1.26 = 12.74
56) the pH of buffer is calcualted by Hendersen equation, which is
pH = pKa + log [salt] / [acid]
[salt] = 0.25
[acid] =0.45
pKa = -logKa = -log1.75 X 10^-5 = 4.75
pH = 4.75 + log [0.25]/[0.45] = 4.49