What is the reduction potential in Volts at 25degree C for the half reaction. 57

Question

What is the reduction potential in Volts at 25degree C for the half reaction. 57. An electrochemical cell is shown on. Which term describes the part of the cell to which the large arrow points? (A) anode (B) cathode (C) electroyne (D) salt bridge 58. The purpose of a salt bridge in a galvanie cell is to (A) maintain charge balance in the cell. (B) correct any volume changes in the cell. (C) provide ions to be oxidized and reduced. (D) provide a path for the flow of free electrons. 59. Given these half reactions Cs^+ + c rightarrow Cs Edegree = -3.026 V F_2 + 2_e rightarrow 2F Edegree 2.866 V what is the standard potential in volts for the reaction 2Cs + F_2 rightarrow 2CsF (A) -5.389 (B) 1,4 (C) +2, +4 (D) +7, +2 Which correctly distriguishes a voltaic cell from an electrolyic cell? 62. Which can act as both an oxidizing agent and a reducing agent? (A) H_2O_2 (B) NaH (C) Na_2O (D) O_3 Which is the strongest reducing agent? (A) Ce (B) Li (C) Li^+ (D) Ce^3- 64. which type of radiation can be blocked by a thin piece of paper (A) I only (B) II only (C) III only (D) I and II 65. A 40.0 gram sample of^235U undergoes decay with a half-life of 7.0times10^3 years. what mass in grams of^235U remains after 2.1times10^9 years? (A) 13. (B) 10. (C) 5.0 (D) 1.4times10^-8 66. Identify the missing reactant in the synthesis of seaborgium, Sg. (A) 24998Cf (B) 24698 Cf (C) 24598Cf (D) 249102No 67. You just spilled 3 mL of concentrated sulfuric acid on your arm. what should you do first? (A) cover with a thick cloth (B) cover with solid NaHCO_3 (C) rinse with 6 M NaOH solution (D) rinse with copious amounts of water 68. which laboratory waste can be discarded down the drain? (A) I only (B) III only (C) II and III only (D) none of the chemicals listed 69. which will have the least vigorous reaction with water? (A) Na (B) Mg (C) K (D) Ca 70. How do the melting points change with increasing atomic number for the diatomic elements in Group 17? (A) decrease (B) increase (C) not predictable (D) little or no change

Explanation / Answer

56) Given reduction half-cell reaction is,

Sn4+ (aq.) + 2e– -------> Sn2+ (aq.)

According to the Nernst equation at equilibrium at 25 oC = 298 K,

Ecell = E0cell – 0.0591/n log ([Products]/[Reactants])

For given cell reaction,

Ecell = E0cell – 0.0591/n log ([Sn2+]/[Sn4+])

We have,

n = Number of electrons transfer = 2

[Sn2+] = 1.0 x 10–6 M and [Sn4+] = 2.00 M and E0cell = 0.15 V

Let us these values in above Nernst equation and calculate E0cell = ?

Ecell = 0.15 – (0.0591/2) log (1.0 x 10–6/2.00)

Ecell = 0.15 – 0.02955 x ( –6.30)

Ecell = 0.15 + 0.19

Ecell = + 0.34 V

Reduction potential for the given cell is +0.34 V Option (C).

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57)

The electrode plate pointed by large arrow is an anode.

In an electrochemical cell electrons leave the anode. Electrons are ionized by the species which is oxidized at an anode and these electrons only leaves the anode.

Hence answer is (A) Anode.

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58)

In galvanic cell salt bridge is to maintain charge balance in the cell. Answer-(A).

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59)

From given data its clear that,

Standard reduction potential of Cs is –3.026 i.e. –ve and hence it has more tendency to act as reducing agent and undergo oxidation self.

Hence Oxidation occur at Cs electrode and hence its an anode and ultimately reduction will occur at F2 electrode side and it’s a cathode.

Standard potential of cell is given as,

Ecell = Standard reduction potential at cathode – Standard reduction potential at an anode.

Ecell = SRP of F2 – SRP of Cs

Ecell = (2.866) – (–3.026)

Ecell = 2.866 + 3.026

Ecell = +5.892 V

Answer option-(C)

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60)

In permanganate ion MnO4– let O.N of Mn be ‘x’

O.N. of O in its metallic oxides is –2

Net charge on MnO4 ion = –1

We can write

Net charge on MnO4 ion = O.N. of Mn + 4 x (O.N. of O) = –1

So, x + 4 x (–2) = –1

x – 8 = –1

x = –1 + 8

x = +7

O.N. of Mn in MnO4– ion is +7

which changes to +2 in Mn2+

Hence,

In given reaction, O.N. of Manganese ion changes from +7 to +2. Answer option-(D).

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61)

Voltaic cell: In voltaic or electrochemical cell a chemical reaction occurs on its own accord which generates an electricity i.e. chemical energy converted into the electrical energy. Hence reactions taking place in voltaic cell are spontaneous and hence G = –ve i.e. G < 0.

Electrolytic cell: An external energy i.e. electricity needed to be provided and then reaction occurs it means in reactions are non-spontaneous and therefore G = +ve i.e. G > 0.

Sign of G best explains voltaic and electrochemical cells.

Answer- (A)

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