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1900 United States presidential election in Rhode Island

The 1900 United States presidential election in Rhode Island took place on November 6, 1900, as part of the 1900 United States presidential election. Voters chose four representatives, or electors to the Electoral College, who voted for president and vice president.


The 1900 United States presidential election in Rhode Island took place on November 6, 1900, as part of the 1900 United States presidential election. Voters chose four representatives, or electors to the Electoral College, who voted for president and vice president.

Rhode Island overwhelmingly voted for the Republican nominee, President William McKinley, over the Democratic nominee, former U.S. Representative and 1896 Democratic presidential nominee William Jennings Bryan. McKinley won Rhode Island by a margin of 24.7% in this rematch of the 1896 presidential election. The return of economic prosperity and recent victory in the Spanish–American War helped McKinley to score a decisive victory.

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

1900 United States presidential election in Rhode Island
RepublicanWilliam McKinley of Ohio (incumbent)Theodore Roosevelt of New York33,78459.74%4100.00%
DemocraticWilliam Jennings Bryan of NebraskaAdlai Ewing Stevenson I of Illinois19,81235.04%00.00%
ProhibitionJohn Granville Woolley of IllinoisHenry Brewer Metcalf of Rhode Island1,5292.70%00.00%
Socialist LaborJoseph F. Malloney of MassachusettsValentine Remmel of Pennsylvania1,4232.52%00.00%
Total56,198100.00%4100.00%
  • United States presidential elections in Rhode Island
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