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1896 United States presidential election in Rhode Island

← 1892 November 3, 1896 1900 →

Nominee William McKinley William Jennings Bryan Party Republican Democratic Home state Ohio Nebraska Running mate Garret Hobart Arthur Sewall Electoral vote 4 0 Popular vote 37,437 14,459 Percentage 68.33% 26.39%

County Results McKinley   60-70%   70-80%

President before election Grover Cleveland Democratic Elected President William McKinley Republican

Elections in Rhode Island
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The 1896 United States presidential election in Rhode Island took place on November 3, 1896, as part of the 1896 United States presidential election. Voters chose four representatives, or electors to the Electoral College, who voted for president and vice president.

Rhode Island voted for the Republican nominee, former governor of Ohio William McKinley, over the Democratic nominee, former U.S. Representative from Nebraska William Jennings Bryan. McKinley won the state by a wide margin of 41.94%.

Bryan, running on a platform of free silver, appealed strongly to Western miners and farmers in the 1896 election, but had little appeal in Northeastern states like Rhode Island.

With 68.33% of the popular vote, Rhode Island would be McKinley's fourth strongest victory in terms of percentage in the popular vote after Vermont, neighboring Massachusetts and New Hampshire.^[1]

Bryan would lose Rhode Island to McKinley again four years later and would later lose the state again in 1908 to William Howard Taft.

• United States presidential elections in Rhode Island