Electrochemistry · Question 6

An electrochemical cell is constructed using half cells in the direction of spontaneous change: Fe(OH)2(s) + 2e- Fe(s) + 2OH-(aq) E = -0.88 V and AgBr(s) + e- Ag(s) + Br-(aq) E = +0.07 V Which of the following options is correct?

1. Overall reaction Fe(s) + 2OH-(aq) + 2AgBr(s) Fe(OH)2(s) + 2Ag(s) + 2Br-(aq)
2. E cell = -0.95 V
3. Fe is reduced in the electrochemical cell
4. E cell is an extensive property

Consider the following two half-cell reactions along with the standard reduction potential given : CO2 + 6H+ + 6e- CH3OH + H2O E red = 0.02 V 12O2 + 2H+ + 2e- H2O E red = 1.23 V A fuel cell was set up using the above two reactions such that the cell operates under the standard condition of 1 bar pressure and 298 K temperature. The fuel cell works with 80% efficiency. If the work derived from the cell using 1 mol of CH3OH is used to compress an ideal gas isothermally against a constant pressure of 1 kPa, then the change in the volume of the gas, V = m3. (nearest integer) Given : F = 96500 C mol-1

Consider the following redox reaction taking place in acidic medium BH4-(aq) + ClO3-(aq) H2BO3-(aq) + Cl-(aq) If the Nernst equation for the above balanced reaction is Ecell = E cell - RTnF Q, then the value of n is . (Nearest integer)

A volume of x mL of 5 M NaHCO3 solution was mixed with 10 mL of 2 M H2CO3 solution to make an electrolytic buffer. If the same buffer was used in the following electrochemical cell to record a cell potential of 235.3 mV, then the value of x = mL (nearest integer). Sn(s) | Sn(OH)62- (0.5 M) | HSnO2- (0.05 M) | OH- | Bi2O3(s) | Bi(s) Consider up to one place of decimal for intermediate calculations [ arrayllGiven: & E Sn(OH)62-|HSnO2- = -0.9 V \ & E Bi2O3|Bi = -0.44 V \ & pKa(H2CO3) = 6.11 \ & 2.303RTF = 0.059 V \ & Antilog (1.29) = 19.5 array ]

For strong electrolyte m increases slowly with dilution and can be represented by the equation m = m - Ac1/2 Molar conductivity values of the solutions of strong electrolyte AB at 18°C are given below : array|l|c|c|c|c| c [mol L-1] & 0.04 & 0.09 & 0.16 & 0.25 \ m [S cm2 mol-1] & 96.1 & 95.7 & 95.3 & 94.9 \ array The value of constant A based on the above data [in S cm2 mol-1/(mol/L)1/2] unit is .

Electricity is passed through an acidic solution of Cu2+ till all the Cu2+ was exhausted, leading to the deposition of 300 mg of Cu metal. However, a current of 600 mA was continued to pass through the same solution for another 28 minutes by keeping the total volume of the solution fixed at 200 mL. The total volume of oxygen evolved at STP during the entire process is mL. (Nearest integer) [Given: Cu2+(aq) + 2e- Cu(s); E red = +0.34 V O2(g) + 4H+ + 4e- 2H2O; E red = +1.23 V Molar mass of Cu = 63.54 g mol-1 Molar mass of O2 = 32 g mol-1 Faraday Constant = 96500 C mol-1 Molar volume at STP = 22.4 L]

Molar conductivity of a weak acid HQ of concentration 0.18 M was found to be 130 of the molar conductivity of another weak acid HZ with concentration of 0.02 M. If Q- happened to be equal to Z-, then the difference of the pKa values of the two weak acids (pKa(HQ) - pKa(HZ)) is (Nearest integer). [Given: degree of dissociation ( ) 1 for both weak acids, : limiting molar conductivity of ions]

In the given electrochemical cell, Ag(s)|AgCl(s)|FeCl2(aq), FeCl3(aq) | Pt(s) at 298 K, the cell potential ( Ecell) will increase when : Choose the correct answer from the options given below :

1. C, D and E Only
2. A and B Only
3. B Only
4. A and E Only

[figure] Consider the above electrochemical cell where a metal electrode (M) is undergoing redox reaction by forming M+ (M M+ + e-). The cation M+ is present in two different concentrations c1 and c2 as shown above. Which of the following statement is correct for generating a positive cell potential?

1. If c1 is present at anode, then c1 > c2.
2. If c1 is present at cathode, then c1 > c2.
3. If c1 is present at cathode, then c1 < c2.
4. If c1 is present at anode, then c1 = c2.

Consider the following electrochemical cell at 298 K Pt | HSnO2-(aq) | Sn(OH)62-(aq) | OH-(aq) | Bi2O3(s) | Bi(s). If the reaction quotient at a given time is 106, then the cell EMF ( Ecell) is 10-1 V (Nearest integer). Given the standard half-cell reduction potential as E Bi2O3/Bi,OH- = -0.44 V and E Sn(OH)62-/HSnO2-,OH- = -0.90 V