Calculate the EMF between copper and silver Ag+e-E=0.89vCu=E=0.34v

Ответ:

jorgepas66

29.01.2022

Depending on the $E^\circ$ value of $\rm Ag^{+} + e^{-} \to Ag\; (s)$ , the cell potential would be:

$0.55\; \rm V$ , using data from this particular question; orapproximately $0.46\; \rm V$ , using data from the CRC handbooks.

Explanation:

In this galvanic cell, the following two reactions are going on:

The conversion between $\rm Ag\; (s)$ and $\rm Ag^{+}$ ions, $\rm Ag^{+} + e^{-} \rightleftharpoons Ag\; (s)$ , andThe conversion between $\rm Cu\; (s)$ and $\rm Cu^{2+}$ ions, $\rm Cu^{2+}\; (aq) + 2\, e^{-} \rightleftharpoons \rm Cu\; (s)$ .

Note that the standard reduction potential of $\rm Ag^{+}$ ions to $\rm Ag\; (s)$ is higher than that of $\rm Cu^{2+}$ ions to $\rm Cu\; (s)$ . Alternatively, consider the fact that in the metal activity series, copper is more reactive than silver. Either way, the reaction is this cell will be spontaneous (and will generate a positive EMF) only if $\rm Ag^{+}$ ions are reduced while $\rm Cu\; (s)$ is oxidized.

Therefore:

The reduction reaction at the cathode will be: $\rm Ag^{+} + e^{-} \to Ag\; (s)$ . The standard cell potential of this reaction (according to this question) is $E(\text{cathode}) = 0.89\; \rm V$ . According to the 2012 CRC handbook, that value will be approximately $0.79\; \rm V$ .The oxidation at the anode will be: $\rm Cu\; (s) \to \rm Cu^{2+} + 2\, e^{-}$ . According to this question, this reaction in the opposite direction ( $\rm Cu^{2+}\; (aq) + 2\, e^{-} \rightleftharpoons \rm Cu\; (s)$ ) has an electrode potential of $0.34\; \rm V$ . When that reaction is inverted, the electrode potential will also be inverted. Therefore, $E(\text{anode}) = -0.34\; \rm V$ .

The cell potential is the sum of the electrode potentials at the cathode and at the anode:

$\begin{aligned}E(\text{cell}) &= E(\text{cathode}) + E(\text{anode}) \\ &= 0.89 \; \rm V + (-0.34\; \rm V) = 0.55\; \rm V\end{aligned}$ .

Using data from the 1985 and 2012 CRC Handbook:

$\begin{aligned}E(\text{cell}) &= E(\text{cathode}) + E(\text{anode}) \\ &\approx 0.7996 \; \rm V + (-0.337\; \rm V) \approx 0.46\; \rm V\end{aligned}$ .

0,0(0 оценок)

Ответ:

alize123213

15.05.2021

Chemistry

Formic acid is the stronger acid.

Explanation:

The strength of an acid is determined by the power of it to dissociate and produce H⁺ ions. Greater the dissociation of an acid, the stronger that acid is. Also, extent of dissociation for particular acid is given by its dissociation constant i.e. Ka.

The greater the Ka value, the stronger an acid is and vice versa.

Calculating Ka value of Pivalic acid:

As,

pKa = -log [Ka]

So,

Ka = 10^-pKa

Putting value of pKa,

Ka = 10⁻⁵

Ka = 1.0 × 10⁻⁵

Calculating Ka value of Formic acid:

As,

pKa = -log [Ka]

So,

Ka = 10^-pKa

Putting value of pKa,

Ka = 10⁻³'⁸

Ka = 1.58 × 10⁻⁴

Conclusion:

As the Ka value of Formic acid is greater than Pivalic acid therefore, it is stronger acid than Pivalic acid.

TIP:

When comparing strength of acids, the acid with the lowest pKa value is considered the strongest acid and vice versa.

0,0(0 оценок)

Calculate the EMF between copper and silver Ag+e-E=0.89v
Cu=E=0.34v

Ответ:

Ответ:

Ask your question to the AI advisor

More tips

Answers on questions: Chemistry

Ask an AI advisor a question