Electrochemistry

One half cell in a voltaic cell is constructed by dipping silver rod in AgNO3 solution of unknown concentration, other half cell is Zn rod dipped in 1 molar solution of ZnSO4. A voltage of 1.60 V is measured at 298 K for this cell. What is the concentration of Ag+ ions used in terms of log x (x = [Ag+])? E Zn2+/Zn = −0.76 V, E Ag+/Ag = +0.80 V, 2.303RTF = 0.059 V

1. 23.9
2. 45.9
3. 2.92
4. 5.94

An electrochemical cell consists of the following two redox couples, Mx+(aq)/M(s) [E red = +0.15 V] and Fe3+(aq)/Fe(s) [E red = -0.036 V]. The cell EMF ( Ecell) is recorded to be 0.2057 V. If the reaction quotient of the electrochemical reaction is found to be 10-2, then the value of x is . (Nearest integer) [Given : M is a p-block metal and 2.303RTF = 0.059 V]

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 ]

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

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

Consider the following electrochemical cell : Pt | O2(g)(1 bar) | HCl(aq) || M2+(aq, 1.0 M) | M(s) The pH above which, oxygen gas would start to evolve at anode is (nearest integer). [Given : . arrayl E M2+/M = 0.994 V \ E O2/H2O = 1.23 V array \ standard reduction potential and RTF(2.303) = 0.059 V at the given condition]

MX is a sparingly soluble salt that follows the given solubility equilibrium at 298 K. MX(s) M+(aq) + X-(aq); Ksp = 10-10 If the standard reduction potential for M+(aq) + e- M(s) is ( E M+/M ) = 0.79 V, then the value of the standard reduction potential for the metal/metal insoluble salt electrode E X-/MX(s)/M is mV. (nearest integer) [Given: 2.303RTF = 0.059 V]

Consider the following half cell reaction Cr2O72-(aq) + 6e- + 14H+(aq) 2Cr3+(aq) + 7H2O( ) The reaction was conducted with the ratio of [ Cr3+]2[ Cr2O72-] = 10-6. The pH value at which the EMF of the half cell will become zero is . (nearest integer value) [Given: standard half cell reduction potential E Cr2O72-,H+/Cr3+ = 1.33 V, 2.303RTF = 0.059 V.]

1 Faraday of electricity was passed through Cu2+(1.5 M, 1 L)/Cu and 0.1 Faraday was passed through Ag+(0.2 M, 1 L)/Ag electrolytic cells. After this the two cells were connected as shown below to make an electrochemical cell. The emf of the cell thus formed at 298 K is mV (nearest integer) [figure] Given : E Cu2+/Cu = 0.34 V E Ag+/Ag = 0.8 V 2.303RTF = 0.06 V

Consider the following electrochemical cell at standard condition. Au(s) | QH2, Q | NH4X(0.01 M) || Ag+(1 M) | Ag(s) Ecell = +0.4 V The couple QH2/Q represents quinhydrone electrode, the half cell reaction is given below: [figure] [Given : E0 Ag+/Ag = +0.8 V and 2.303RTF = 0.06 V] The pKb value of the ammonium halide salt (NH4X) used here is . (nearest integer)

For a Mg | Mg2+(aq) || Ag+(aq) | Ag the correct Nernst Equation is :

1. Ecell = E cell - RT2F [Ag+]2[Mg2+]
2. Ecell = E cell - RT2F [Mg2+][Ag+]
3. Ecell = E cell - RT2F [Ag+][Mg2+]
4. Ecell = E cell + RT2F [Ag+]2[Mg2+]