in a galvonic cell 1) reduction occur at anode 2) oxidation occur at cathode 3) electrical energy produces chemical energy 4) chemical reaction produce electrical energy
Answers
Answer:
see below
Explanation:
For electrolytic cell following are true statements:
(a) Nonspontaneous Reduction.
(b) electrical energy → chemical energy
(c) Oxidation at anode (+ve charged electrode)
Reduction at cathode (−ve charged electrode)
I think it will help
Answer:
are driven by a spontaneous chemical reaction that produces an electric current through an outside circuit. These cells are important because they are the basis for the batteries that fuel modern society. But they are not the only kind of electrochemical cell. The reverse reaction in each case is non-spontaneous and requires electrical energy to occur.
Introduction
The general form of the reaction can be written as:
Reactants⇌Products+Electrical EnergySpontaneous⟶⟵Non spontaneous(1)
It is possible to construct a cell that does work on a chemical system by driving an electric current through the system. These cells are called electrolytic cells. Electrolytic cells, like galvanic cells, are composed of two half-cells--one is a reduction half-cell, the other is an oxidation half-cell. The direction of electron flow in electrolytic cells, however, may be reversed from the direction of spontaneous electron flow in galvanic cells, but the definition of both cathode and anode remain the same, where reduction takes place at the cathode and oxidation occurs at the anode. Because the directions of both half-reactions have been reversed, the sign, but not the magnitude, of the cell potential has been reversed.
Electrolytic cells are very similar to voltaic (galvanic) cells in the sense that both require a salt bridge, both have a cathode and anode side, and both have a consistent flow of electrons from the anode to the cathode. However, there are also striking differences between the two cells. The main differences are outlined below:
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Figure 1 : Electrochemical Cells. A galvanic cell (left) transforms the energy released by a spontaneous redox reaction into electrical energy that can be used to perform work. The oxidative and reductive half-reactions usually occur in separate compartments that are connected by an external electrical circuit; in addition, a second connection that allows ions to flow between the compartments (shown here as a vertical dashed line to represent a porous barrier) is necessary to maintain electrical neutrality. The potential difference between the electrodes (voltage) causes electrons to flow from the reductant to the oxidant through the external circuit, generating an electric current. In an electrolytic cell (right), an external source of electrical energy is used to generate a potential difference between the electrodes that forces electrons to flow, driving a nonspontaneous redox reaction; only a single compartment is employed in most applications. In both kinds of electrochemical cells, the anode is the electrode at which the oxidation half-reaction occurs, and the cathode is the electrode at which the reduction half-reaction occurs.
Table 1 : Properties of Galvanic and Electrochemical of [PtCl6]2-, using an average current of 10 amperes at an 80% electrode efficiency?
8.4
5.4
16.8
11.2
12.4
8) How many faradays are required to reduce 1.00 g of aluminum(III) to the aluminum metal?
1.00
1.50
3.00
0.111
0.250
9) Find the standard cell potential for an electrochemical cell with the following cell reaction.
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
Answers
1). Cl- chlorine H+ hydrogen
Cl- chlorine Cu2+ copper
I- iodine H+ Hhydrogen
2) 12 mol e– is required to plate 2 mol Cr, giving us a stoichiometric ratio S(e–/Cr). Then the Faraday constant can be used to find the quantity of charge.
nCr \xrightarrow{S\text{(}e^{-}\text{/Cr)}} ne– \xrightarrow{F} Q
Q = 1.386 mol Cr × \frac{\text{12 mol }e^{-}}{\text{2 mol Cr}} × \frac{\text{9}\text{.649 }\times \text{ 10}^{\text{4}}\text{ C}}{\text{ 1 mol }e^{-}} = 8.024 × 105 C
3) The product of current and time gives us the quantity of electricity, Q. Knowing this we easily calculate the amount of electrons, ne–. From the first half-equation we can then find the amount of peroxydisulfuric acid, and the second leads to nH2O2 and finally to mH2O2.
m_{\text{H}_{\text{2}}\text{O}_{\text{2}}}=\text{0}\text{.893 A }\times \text{ 3600 s }\times \text{ }\frac{\text{1 mol }e^{-}}{\text{96 490 C}}\text{ }\times \text{ }\frac{\text{1 mol H}_{\text{2}}\text{S}_{\text{2}}\text{O}_{\text{8}}}{\text{2 mol }e^{-}}
=\frac{\text{1 mol H}_{\text{2}}\text{O}_{\text{2}}}{\text{1 mol H}_{\text{2}}\text{S}_{\text{2}}\text{O}_{\text{8}}}\text{ }\times \text{ }\frac{\text{34}\text{.01 g H}_{\text{2}}\text{O}_{\text{2}}}{\text{1 mol H}_{\text{2}}\text{O}_{\text{2}}}
= 05666 \frac{\text{A s}}{\text{C}} × g H2O2 = 0.5666 g H2O2
4) 0.259 mol e-
5) d
6) d
7) b
8) d
9) Write the half-reactions for each process.
Zn(s) → Zn2+(aq) + 2 e-
Cu2+(aq) + 2 e- → Cu(s)
Look up the standard potentials for the reduction half-reaction.
Eoreduction of Cu2+ = + 0.339 V
Eoreduction of Zn2+ = - 0.762 V
Determine the overall standard cell potential.
Eocell = + 1.101 V