---
title: "A student investigates the standard thermodynamic and electrochemical properties of the reaction represented by the following equation at \\(298\\text{ K}\\):  \\[ \\text{Sn(s)} + 2\\,\\text{Ag}^+\\text{(aq)} \\rightleftharpoons \\text{Sn}^{2+}\\text{(aq)} + 2\\,\\text{Ag(s)} \\quad E^\\circ_{\\text{cell}} = +0.94\\text{ V} \\]  Which of the following correctly identifies the sign of \\(\\Delta G^\\circ\\) and the magnitude of \\(K\\) relative to \\(1\\), along with the appropriate justification?"
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url: "https://nerd-notes.com/ubq/119959/"
date_modified: "2026-08-21T08:30:38+00:00"
---

# A student investigates the standard thermodynamic and electrochemical properties of the reaction represented by the following equation at \(298\text{ K}\):

\[ \text{Sn(s)} + 2\,\text{Ag}^+\text{(aq)} \rightleftharpoons \text{Sn}^{2+}\text{(aq)} + 2\,\text{Ag(s)} \quad E^\circ_{\text{cell}} = +0.94\text{ V} \]

Which of the following correctly identifies the sign of \(\Delta G^\circ\) and the magnitude of \(K\) relative to \(1\), along with the appropriate justification?

A student investigates the standard thermodynamic and electrochemical properties of the reaction represented by the following equation at \(298\text{ K}\):

\[ \text{Sn(s)} + 2\,\text{Ag}^+\text{(aq)} \rightleftharpoons \text{Sn}^{2+}\text{(aq)} + 2\,\text{Ag(s)} \quad E^\circ_{\text{cell}} = +0.94\text{ V} \]

Which of the following correctly identifies the sign of \(\Delta G^\circ\) and the magnitude of \(K\) relative to \(1\), along with the appropriate justification?

- **A.** \(\Delta G^\circ > 0\) and \(K < 1\), because a positive value of \(E^\circ_{\text{cell}}\) indicates that electrical work must be absorbed from the surroundings to drive the transfer of electrons.
- **B.** \(\Delta G^\circ < 0\) and \(K < 1\), because according to \(\Delta G^\circ = -RT\ln K\), a negative value of \(\Delta G^\circ\) requires \(\ln K\) to be negative, which corresponds to \(0 < K < 1\).
- **C.** \(\Delta G^\circ < 0\) and \(K > 1\), because \(E^\circ_{\text{cell}} > 0\) yields a negative \(\Delta G^\circ\) through \(\Delta G^\circ = -nFE^\circ_{\text{cell}}\), which requires \(\ln K > 0\) and thus \(K > 1\) in \(\Delta G^\circ = -RT\ln K\).
- **D.** \(\Delta G^\circ < 0\) and \(K > 1\), because the positive cell voltage ensures that the rate of the forward reaction is substantially faster than the rate of the reverse reaction, driving the reaction to irreversible completion.

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/119959/*