If a piece of copper metal is dipped into a solution containing Cr3+ ions, what will happen? Explain with Ev values
DrBob... Could you explain how you got negative...I thought E = reduced - oxidized E = 0.34 V - (-0.74 V) = 1.08 V
Nothing happens.
Cr^3+ + 3e ==> -0.74v
Cu ==> Cu^2+ + 2e -0.337
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2Cr^3+ + 3Cu ==> 3Cu^2+ + 2Cr E = -1.077v
The voltage is negative, hence the reaction, as written, will not occur. It will occur in the reverse direction.
YOU JUST HAVE TO FOLLOW THE RULES & ABIDE WITH IT
To understand what will happen when a piece of copper metal is dipped into a solution containing Cr3+ ions, we need to refer to the standard reduction potentials, commonly known as E° values.
E° values provide a measure of the tendency of a species to gain or lose electrons. A positive E° value indicates a species that is more likely to be reduced (gain electrons), while a negative E° value indicates a species that is more likely to be oxidized (lose electrons).
For our case, we need to compare the E° values of copper and chromium ions to determine which one is more likely to undergo reduction and which one is more likely to undergo oxidation.
The standard reduction potential for copper (Cu2+ + 2e- → Cu) is +0.34 V.
The standard reduction potential for chromium (Cr3+ + 3e- → Cr) is -0.74 V.
Comparing these values, we can see that the E° value for copper is higher than that of chromium. This means that copper has a greater tendency to be reduced and gain electrons, while chromium has a greater tendency to be oxidized and lose electrons.
Therefore, when a piece of copper metal is dipped into a solution containing Cr3+ ions, the copper metal will undergo oxidation (lose electrons) to form Cu2+ ions. At the same time, the Cr3+ ions in the solution will be reduced (gain electrons) to chromium metal.
In practical terms, this means that the copper metal will slowly corrode (oxidation) while the Cr3+ ions in the solution will be reduced, resulting in the formation of chromium metal.