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3 years ago

5

What is the voltage for the nonspontaneous reaction between silver (Ag) and copper (Cu) and their ions?

1 answer:

gayaneshka [121]3 years ago

5 0

For an non spontaneous reaction between silver (Ag) and copper (Cu) and their ions, Cu is the oxidizing agent while Ag+ is the reducing agent,
The following reactions will take place;
Anode Cu = Cu+2 + 2e- E= +0.34 volts
Cathode; Ag+ + e = Ag E = +0.80 volts
The net reaction will be Cu + 2Ag+ = Cu+2 + 2Ag
Thus, the voltage will be
= +0.80 - (+0.34)

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emmasim [6.3K]

Answer:

$q_{rxn}=-488.86\frac{kJ}{mol}$

Explanation:

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In this case, since this calorimetry problems can by analyzed by figuring out that the heat lost due to the combustion of C6H12O6 is gained by the water and the calorimeter which undergo the mentioned temperature rise, we can write:

$Q_{rxn}=-(m_wC_w+C_c)(T_2-T_1)$

Thus, by plugging in the given data, we obtain:

$Q_{rxn}=-(950.0g*4.184\frac{J}{g\°C} +933\frac{J}{\°C})(42.3\°C-35.0\°C)\\\\Q_{rxn}=-17.5kJ$

Next, we compute the moles of C6H12O6 by using its molar mass (180.18 g/mol) as shown below:

$n=6.45g*\frac{1mol}{180.18g} =0.0358mol$

Thus, the value for q of the reaction turns out:

$q_{rxn}=\frac{Q_{rxn}}{n} \\\\q_{rxn}=\frac{17.5kJ}{0.0358mol} \\\\q_{rxn}=-488.86\frac{kJ}{mol}$

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