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The electrode with reaction :Cr(2)O(7)^(...

The electrode with reaction `:Cr_(2)O_(7)^(2-)(aq)+14H^(o+)(aq)+6e^(-) rarr 2Cr^(3+)(aq)+7H_(2)O` can be represented as

A

`Pt|H^(o+)(aq),Cr_(2)O_(7)^(2-)(aq)`

B

`Pt|H^(o+)(aq),Cr_(2)O_(7)^(2-)(aq),Cr^(3+)(aq)`

C

`Pt_(H2)|H^(o+)(aq),Cr_(2)O_(7)^(2-)`

D

`Pt_(H2)|H^(o+)(aq),Cr_(2)O_(7)^(2-)(aq),Cr^(3+)(aq)`

Text Solution

AI Generated Solution

The correct Answer is:
To represent the electrode reaction given by the equation: \[ \text{Cr}_2\text{O}_7^{2-}(aq) + 14 \text{H}^+(aq) + 6 e^- \rightarrow 2 \text{Cr}^{3+}(aq) + 7 \text{H}_2\text{O} \] we need to identify the components involved in the half-reaction and determine how to represent this electrochemical cell. ### Step-by-Step Solution: 1. **Identify the Reduction Reaction**: - The provided reaction shows that the dichromate ion (\(\text{Cr}_2\text{O}_7^{2-}\)) is being reduced to chromium ions (\(\text{Cr}^{3+}\)). This indicates that the chromium is gaining electrons (6 electrons in this case). 2. **Determine the Oxidation State**: - In the dichromate ion, chromium has an oxidation state of +6, while in the chromium ion (\(\text{Cr}^{3+}\)), it has an oxidation state of +3. This confirms that the chromium is being reduced. 3. **Identify the Cathode**: - Since reduction occurs at the cathode, the half-reaction represents the cathode process where dichromate ions are reduced to chromium ions. 4. **Identify the Anode Reaction**: - In an electrochemical cell, oxidation occurs at the anode. In this case, hydrogen gas (\(H_2\)) can be oxidized to \(H^+\) ions. Therefore, the anode reaction can be represented as: \[ \text{H}_2(g) \rightarrow 2 \text{H}^+(aq) + 2 e^- \] 5. **Construct the Cell Representation**: - The complete cell representation combines the anode and cathode reactions. The standard representation of an electrochemical cell is: \[ \text{Anode} \, | \, \text{Electrolyte} \, || \, \text{Electrolyte} \, | \, \text{Cathode} \] - For our case, the anode is where hydrogen gas is oxidized, and the cathode is where dichromate is reduced. Therefore, the cell can be represented as: \[ \text{Pt} \, | \, \text{H}_2(g) \, | \, \text{H}^+(aq) \, || \, \text{Cr}_2\text{O}_7^{2-}(aq), \text{Cr}^{3+}(aq) \, | \, \text{H}_2\text{O} \] 6. **Select the Correct Option**: - Based on the above analysis, the correct representation of the electrochemical cell is option D, which includes the platinum electrode, hydrogen gas, and the dichromate ions. ### Final Answer: The electrode can be represented as: \[ \text{Pt} \, | \, \text{H}_2(g) \, | \, \text{H}^+(aq) \, || \, \text{Cr}_2\text{O}_7^{2-}(aq), \text{Cr}^{3+}(aq) \, | \, \text{H}_2\text{O} \]

To represent the electrode reaction given by the equation: \[ \text{Cr}_2\text{O}_7^{2-}(aq) + 14 \text{H}^+(aq) + 6 e^- \rightarrow 2 \text{Cr}^{3+}(aq) + 7 \text{H}_2\text{O} \] we need to identify the components involved in the half-reaction and determine how to represent this electrochemical cell. ...
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What type of a battery is lead storage battery? Write the anode and cathode reactions and the overall cell reaction occurring in the operation of a lead storage battery. (b) Calucluate the potential for half-cell containing. 0.10 M K_(2)Cr_(2)O_(7)(aq), 0.20 M Cr^(3+) (aq) and 1.0 xx 10^(-4) M H^(+) (aq) The half -cell reaction is Cr_(2)O_(7)^(2-) (aq) + 4H^(+) (aq) + 6 e^(-) to 2 Cr^(3+) (aq) + 7H_(2)O(l) and the standard electron potential is given as E^(o) = 1.33V .

The cell reaction Cr_2O_7^(2-)(aq)+14H^+(aq)+6Fe^(2+)(aq)to6Fe^(3+)(aq)+2Cr^(3+)(aq)+7H_2O(l) is best represented by:

In the reaction: Cr_(2)O_(7)^(2-) + 14H^(o+) + 6I^(Ө) rarr2Cr^(3+) + 3H_(2)O + 3I_(2) Which element is reduced?

For reaction : Cr_(2)O_(7)^(-2)+14H^(+)rarr2Cr^(+3)+7H_(2)O , How many e^(-)s are required

Cr_(2)O_(7)^(2-)+H^(+)+SO_(3)^(2-) to Cr^(3+)(aq.)+SO_(4)^(2-)

Cr_(2)O_(7)^(2-)+H^(+)+SO_(3)^(2-) to Cr^(3+)(aq.)+SO_(4)^(2-)

Consider the reaction : Cr_(2)O_(7)^(2-)+14H^(o+)+6e^(-) rarr 2Cr^(+3)+8H_(2)O What is the quantity of electricity in coulombs needed to reduce 1 mol e of Cr_(2)O_(7)^(2-) ?

Calculate the potential for half-cell containing 0.10 MK_(2)Cr_(2)O_(7)(aq),0.20M Cr^(3+)(aq) and 1.0xx10^(-4)MH^(+)(aq) . The half-cell reaction is Cr_(2) O_(7)^(2-)(aq) +14H^(+) +6e^(-) to 2Cr^(3+)(aq) +7H_(2)O(l)

Standard electrode potential data are useful for understanding the suitability of an oxidant in a redox titration. Some half cell reaction and their standard potentials are given below: MnO_(4)^(-)(aq) +8H^(+)(aq) +5e^(-) rarr Mn^(2+)(aq) +4H_(2)O(l) E^(@) = 1.51V Cr_(2)O_(7)^(2-)(aq) +14H^(+) (aq) +6e^(-) rarr 2Cr^(3+)(aq) +7H_(2)O(l), E^(@) = 1.38V Fe^(3+) (aq) +e^(-) rarr Fe^(2+) (aq), E^(@) = 0.77V CI_(2)(g) +2e^(-) rarr 2CI^(-)(aq), E^(@) = 1.40V Identify the only correct statement regarding quantitative estimation of aqueous Fe(NO_(3))_(2)

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