Consider the half reaction,
2H+(aq) + 2e– → H2(g); E° = 0.0V
Au3+(aq) + 3e– → Au(s); E° = 1.50V
Since E° (1.50V) for (Au3+/Au) is higher than that H+/-\(\frac{1}{2}\)H2 (0.0V), therefore, Au3+ can be more easily reduced than H+ ions Au3+ can be reduced to Au metal by H2 but H+ cannot oxidize metallic gold to Au3+ ions. In other words, metallic gold does not dissolve in 1M HCl instead H2 gas can reduce gold salt to metallic gold.